Consider the digital sum function in base 2. In 1948, R. Bellman and H.N. Shapiro proved a formula for the sum of digital sum functions in base 2 up to a certain number with an error term. However, there was a mistake in their proof. We are able to correct the mistake and walk in their footsteps while retaining their idea and method to prove the theorem. We also generalized their method to prove the same theorem for general base

Gadd, Cora

Wong, Ka Lun

English

California State University (CSU): Open Journal Systems

A Generalization to Bellman and Shapiro’s Methodon the Sum of Digital Sum FunctionsC. Gadd and K.L. WongAbstract - Let α(n) be the sum of digits of n written in base 2. In 1948, R. Bellmanand H.N. Shapiro proved that∑n≤x α(n) =x log x2 log 2 + O(x log log x). However, there was amistake in their proof. We are able to correct the mistake and walk in their footsteps whileretaining their idea and method to prove the theorem. We also generalize their method toprove the same theorem for general base.Keywords : digital sums; base q representationMathematics Subject Classification (2020) : 11A63;1 IntroductionLet q > 1 be a fixed integer and denote by αq(n) the sum of digits of n written in baseq. Let f(x) and g(x) be real-valued functions and g(x) be positive for all large enoughvalues of x. We write f(x) = O(g(x)) if there exists a positive real number M and a realnumber x0 such that |f(x)| ≤Mg(x) for all x ≥ x0. It was L.E. Bush [3] who first showedin 1940 thatAq(x) :=∑n≤xαq(n) ∼q − 12 log qx log x as x→∞. (1)In 1949, L. Mirsky [6] proved thatAq(x) =q − 12 log qx log x+O(x), (2)using a special way to count numbers with a certain representation in base q. It was shownthat O(x) is the best possible error term. A weaker error term, namely O(x log log x), forbase 2 was discovered a year earlier by Bellman and Shapiro [2]. Many other authors haveproved (2) later on using different methods (see [1], [4] and [5]). For example, Ballot in[1] defined a special function Bq(N) which is equal to Aq(N) at a power of q, but differentat other numbers. By bounding the difference Aq(N) − Bq(N), the author was able toshow that Aq(N)−Bq(N) = O(N). After proving an asymptotic formula for Bq(N), (2)was established.This paper stemmed from noticing what seemed — at first — a major mistake inBellman and Shapiro’s proof. The mistake came in the formula [2, (3.10)] with a wrongthe pump journal of undergraduate research 5 (2022), 176–187 176evaluation of A2(N). Here, we evaluate A2(N) properly while, nevertheless, followingtheir proof and idea to obtain the same error term. The corrected formula appears inequation (3) here. We mostly use their notation too. What distinguishes the Bellman–Shapiro method from other approaches is the introduction of a parameter µ(N) for whereto split a sum S into two subsums involved in expressing A2(N). They then choose µ(N)so the two partial subsums have the same order of magnitude. In Section 1, we reviseBellman and Shapiro’s proof and prove the following theorem.Theorem 1.1 We haveA2(x) =x log x2 log 2+O(x log log x).In Section 3, we proved the same theorem for general bases.Theorem 1.2 We haveAq(x) =q − 12 log qx log x+O(x log log x).2 Proof of Theorem 1.1Proof. Let α(n) = α2(n). From the dyadically additive property of α(n), it follows thatA2(2n) =∑1≤k≤2n−1α(k) +∑1≤r<2n−1α(2n−1 + r) + α(2n)= A2(2n−1) +[(2n−1 − 1) + A2(2n−1)− α(2n−1)]+ 1= 2A2(2n−1) + 2n−1 − 1.Iterating this relationship, we obtainA2(2n) = 2(2A2(2n−2) + 2n−2 − 1)+ 2n−1 − 1 = 22A2(2n−2) + 2 · 2n−1 − 3= 22(2A2(2n−3) + 2n−3 − 1)+ 2 · 2n−1 − 3 = 23A2(2n−3) + 3 · 2n−1 − 7= 23(2A2(2n−4) + 2n−4 − 1)+ 3 · 2n−1 − 7 = 24A2(2n−4) + 4 · 2n−1 − 15= . . . = 2nA2(2n−n) + n · 2n−1 − (2n − 1) = n2n−1 + 1.If N = 2n1 + 2n2 + · · ·+ 2nt , with n1 > n2 > · · · > nt, thenA2(N) =∑k≤2n1α(k) +∑2n1<k≤2n1+2n2α(k) + · · ·+∑2n1+2n2+···+2nt−1<k≤2n1+2n2+···+2ntα(k)=∑1≤k≤2n1α(k) +∑1≤k≤2n2α(2n1 + k) +∑1≤k≤2n3α(2n1 + 2n2 + k) + · · ·the pump journal of undergraduate research 5 (2022), 176–187 177= A2(2n1) +(2n2 + A2(2n2))+(2 · 2n3 + A2(2n3))+ · · ·+((t− 1) · 2nt + A2(2nt))=t∑i=1A2(2ni) +t∑i=1(i− 1)2ni =t∑i=1(ni2ni−1 + 1) +t∑i=1(i− 1)2ni=12t∑i=1ni2ni +t∑i=1(i− 1)2ni + t.Thus, the corrected formula for A2(N) isA2(N) =12t∑i=1ni2ni +t∑i=1(i− 1)2ni + t, (3)instead of12t∑i=1ni2ni +N,in the Bellman–Shapiro paper [2].Let S =t∑i=1ni2ni =n1∑j=0aj(n1 − j)2n1−j, where aj = 0 or 1 depending upon thepresence or absence of 2n1−j in the sum N = 2n1 + 2n2 + · · ·+ 2nt . ThenS =µ(N)∑j=0aj(n1 − j)2n1−j +n1∑j=µ(N)+1aj(n1 − j)2n1−jwhere n1 ≥ µ(N) + 1 ≥ 1, and µ(N) will be chosen advantageously below.S = n1µ(N)∑j=0aj2n1−j −µ(N)∑j=0ajj2n1−j +O(n12n1−µ(N))= n1n1∑j=0aj2n1−j − n1n1∑j=µ(N)+1aj2n1−j −µ(N)∑j=0ajj2n1−j +O(n12n1−µ(N))= n1N +O(n12n1−µ(N)) +O(µ(N)2n1) +O(n12n1−µ(N))= n1N +O(µ(N)2n1) +O(n12n1−µ(N)).Choose µ(N) to make the two error terms of comparable size, i.e., such thatn12n1−µ(N) = µ(N)2n1 ,or n12−µ(N) = µ(N).Additionally,N = 2n1 + 2n2 + · · ·+ 2nt = 2n1(1 + 2n2−n1 + · · ·+ 2nt−n1)log2N = n1 + log2(1 + 2n2−n1 + · · ·+ 2nt−n1).Since 1 < 1 + 2n2−n1 + · · ·+ 2nt−n1 < 2, we havethe pump journal of undergraduate research 5 (2022), 176–187 1780 < log2(1 + 2n2−n1 + · · ·+ 2nt−n1) < 1n1 < log2N < n1 + 1.Thus, n1 = blog2Nc =⌊logNlog 2⌋.Therefore,µ(N) =⌊logNlog 2⌋2−µ(N).It follows that µ(N) ∼ log(logN)log 2as N −→∞. Thus,S =N logNlog 2+O(N log logN).Now, let R =t∑i=1(i− 1)2ni + t. To prove the theorem it suffices to show that R = O(N).But we can show more precisely that R ≤ N .A table of N ≤ 32 values, binary representation, R values, and the increase in R valuesis presented below. The table shows some repeating patterns and suggests that R ≤ N .A graph of R versus N is also included below.N Binary R R Increase of R1 1 1 + 0 · 20 12 10 1 + 0 · 21 1 03 11 2 + 0 · 21 + 1 · 20 3 24 100 1 + 0 · 22 1 -25 101 2 + 0 · 22 + 1 · 20 3 26 110 2 + 0 · 22 + 1 · 21 4 17 111 3 + 0 · 22 + 1 · 21 + 2 · 20 7 38 1000 1 + 0 · 23 1 -69 1001 2 + 0 · 23 + 1 · 20 3 210 1010 2 + 0 · 23 + 1 · 21 4 111 1011 3 + 0 · 23 + 1 · 21 + 2 · 20 7 312 1100 2 + 0 · 23 + 1 · 22 6 -113 1101 3 + 0 · 23 + 1 · 22 + 2 · 20 9 314 1110 3 + 0 · 23 + 1 · 22 + 2 · 21 11 215 1111 4 + 0 · 23 + 1 · 22 + 2 · 21 + 3 · 20 15 416 10000 1 + 0 · 24 1 -1417 10001 2 + 0 · 24 + 1 · 20 3 218 10010 2 + 0 · 24 + 1 · 21 4 119 10011 3 + 0 · 24 + 1 · 21 + 2 · 20 7 3the pump journal of undergraduate research 5 (2022), 176–187 17920 10100 2 + 0 · 24 + 1 · 22 6 -121 10101 3 + 0 · 24 + 1 · 22 + 2 · 20 9 322 10110 3 + 0 · 24 + 1 · 22 + 2 · 21 11 223 10111 4 + 0 · 24 + 1 · 22 + 2 · 21 + 3 · 20 15 424 11000 2 + 0 · 24 + 1 · 23 10 -525 11001 3 + 0 · 24 + 1 · 23 + 2 · 20 13 326 11010 3 + 0 · 24 + 1 · 23 + 2 · 21 15 227 11011 4 + 0 · 24 + 1 · 23 + 2 · 21 + 3 · 20 19 428 11100 3 + 0 · 24 + 1 · 23 + 2 · 22 19 029 11101 4 + 0 · 24 + 1 · 23 + 2 · 22 + 3 · 20 23 430 11110 4 + 0 · 24 + 1 · 23 + 2 · 22 + 3 · 21 26 331 11111 5 + 0 · 24 + 1 · 23 + 2 · 22 + 3 · 21 + 4 · 20 31 532 100000 1 + 0 · 25 1 -32Table 1: R vs NFigure 1: R vs Nthe pump journal of undergraduate research 5 (2022), 176–187 180Here we prove R ≤ N . ConsiderN −R =t∑i=12ni −(t∑i=1(i− 1)2ni + t)= 2n1 − (2n3 + 2 · 2n4 + 3 · 2n5 + · · ·+ (t− 2)2nt + t).Note that2n3 + 2n4 + 2n5 + · · ·+ 2nt ≤ 2n2 − 1,2n4 + 2n5 + · · ·+ 2nt ≤ 2n3 − 1,2n5 + · · ·+ 2nt ≤ 2n4 − 1,...2nt−1 + 2nt ≤ 2nt−2 − 1,2nt ≤ 2nt−1 − 1.If we sum up all the inequalities above, we have2n3 + 2 · 2n4 + 3 · 2n5 + · · ·+ (t− 2)2nt ≤ 2n2 + 2n3 + · · ·+ 2nt−1 − (t− 2)2n3 + 2 · 2n4 + 3 · 2n5 + · · ·+ (t− 2)2nt + t ≤ 2n2 + 2n3 + · · ·+ 2nt−1 + 2≤ 2n1 − 1− 2nt + 2≤ 2n1 − (2nt − 1)≤ 2n1 .Therefore, N − R ≥ 0 and R ≤ N . It is interesting to note that we only have anequality when all ni are consecutive and nt = 0. In other words, R = N only whenN = 2n − 1 for some n. This can be seen from the table and the graph above.Now, we have proved the following:A2(N) =N logN2 log 2+O(N log logN).3 Proof of Theorem 1.2Let q ≥ 2 be an arbitrary base. To use the same method as in the base 2 case, we firstderive a formula for Aq(qn). This formula is proved in [1] using a different method. Theauthor used a clever way to pair terms in the sum. Here we use a recursive method toprove the formula.the pump journal of undergraduate research 5 (2022), 176–187 181Lemma 3.1 Let Aq(x) =∑k≤x αq(k) where αq(k) is the sum of digits of k written inbase q. ThenAq(qn) =∑k≤qnαq(k) = 1 +q − 12nqn.Proof. Following the idea from [2], we split the sum first at qn−1. A difference here isthat we also need to run through different coefficients of qn−1.Aq(qn) =∑k≤qnα(k) =∑k≤qn−1α(k) +q−1∑m=1( qn−1−1∑r=1α(m · qn−1 + r) + α((m+ 1)qn−1))=∑k≤qn−1α(k) +q−1∑m=1( qn−1−1∑r=1α(m · qn−1 + r))+q−1∑m=1α((m+ 1)qn−1)=∑k≤qn−1α(k) +q−1∑m=1( qn−1−1∑r=1α(m · qn−1))+q−1∑m=1( qn−1−1∑r=1α(r))+q−1∑m=1α((m+ 1)qn−1)=∑k≤qn−1α(k) +q−1∑m=1m(qn−1 − 1) + (q − 1)qn−1−1∑r=1α(r) +q−1∑m=1m=∑k≤qn−1α(k) +q−1∑m=1m(qn−1 − 1) + (q − 1)qn−1∑r=1α(r) +q−2∑m=1m=qAq(qn−1) + qn−1q−1∑m=1m−q−1∑m=1m+q−2∑m=1m=qAq(qn−1) + qn−1(q(q − 1)2)− (q − 1)=qAq(qn−1) +q − 12qn − (q − 1)=q[qAq(qn−2) +q − 12qn−1 − (q − 1)]+q − 12qn − (q − 1)=q2Aq(qn−2) + 2 · q − 12qn − q(q − 1)− (q − 1)=q3Aq(qn−3) + 3 · q − 12qn − q2(q − 1)− q(q − 1)− (q − 1)...=qnAq(qn−n) + n · q − 12qn − (q − 1)(1 + q + q2 + · · ·+ qn−1)the pump journal of undergraduate research 5 (2022), 176–187 182= qn +q − 12nqn − (qn − 1)= 1 +q − 12nqn.Now let N = c1qn1 + c2qn2 + c3qn3 + · · · + ctqnt , where n1 > n2 > · · · > nt ≥ 0. Notethat in the base q expansion of N , the coefficients are not necessarily 1. Thus, we need aformula for Aq(cnqn) while the base 2 case only needs a formula for A2(2n). Fortunately,when we split the sum Aq(cnqn) at different multiples of qn, we recover many copies ofAq(qn). Thus, we haveAq(cnqn) =∑k≤cnqnα(k)=∑1≤k≤qnα(k) +∑1≤r≤qn−1α(qn + r) + α(2 · qn) + · · ·+∑1≤r≤qn−1α((cn − 1)qn + r)+ α(cnqn)=Aq(qn) + (qn − 1) +∑1≤r≤qn−1α(r) + α(2 · qn) + · · ·+ (cn − 1)(qn − 1)+∑1≤r≤qn−1α(r) + α(cnqn)=cnAq(qn) + qn(1 + 2 + · · ·+ (cn − 1))=cn(1 +q − 12nqn)+ qn(cn(cn − 1)2)=cn +(cn(q − 1)2n+cn(cn − 1)2)qn.This formula generalizes the formula in Lemma 3.1. Now we can look at Aq(N).Similar to the base 2 case, we split the sum Aq(N) at sums of different numbers of thefirst few summands of N . Thus, we haveAq(N) =∑k≤c1qn1α(k) +∑c1qn1<k≤c1qn1+c2qn2α(k) + · · ·+∑c1qn1+c2qn2+···+ct−1qnt−1<k≤c1qn1+c2qn2+···+ctqntα(k)=∑k≤c1qn1α(k) +∑1≤r≤c2qn2α(c1qn1 + r) + · · ·+∑1≤r≤ctqntα(c1qn1 + c2qn2 + · · ·+ ct−1qnt−1 + r)the pump journal of undergraduate research 5 (2022), 176–187 183=∑k≤c1qn1α(k) +∑1≤r≤c2qn2α(c1qn1) +∑1≤r≤c2qn2α(r) + · · ·+∑1≤r≤ctqntα(c1qn1 + c2qn2 + · · ·+ ct−1qnt−1) +∑1≤r≤ctqntα(r)=Aq(c1qn1) + c1 · c2qn2 + Aq(c2qn2) + · · ·+ (c1 + c2 + · · ·+ ct−1)ctqnt + Aq(ctqnt)=t∑i=1Aq(ciqni) +t−1∑i=1(ci+1i∑j=1cj)qni+1=t∑i=1[ci +(ci(q − 1)2ni +ci(ci − 1)2)qn1]+t−1∑i=1(ci+1i∑j=1cj)qni+1=t∑i=1(ci(q − 1)2ni +ci(ci − 1)2)qni +t∑i=1ci +t−1∑i=1(ci+1i∑j=1cj)qni+1=t∑i=1(ci(q − 1)2ni +ci(ci − 1)2)qni +t∑i=1ci +t∑i=2(c1 + c2 + · · ·+ ci−1)ciqni .Similar to the base 2 case, there are two parts in the formula of Aq(N). They bothbehave in the same ways as the two parts in the base 2 case respectively. To show that,we first letSq =t∑i=1(ci(q − 1)2ni +ci(ci − 1)2)qni and Rq =t∑i=1ci +t∑i=2(c1 + c2 + · · ·+ ci−1)ciqni .We will prove the general results, namelySq =q − 12· N logNlog q+O(N log logN) and Rq ≤ N.Then, Theorem 1.2 will follow.Now we proceed the µ(N) approach to Sq. We haveSq =t∑i=1(ci(q − 1)2ni +ci(ci − 1)2)qni =n1∑j=0aj(dj(q − 1)2(n1 − j) +dj(dj − 1)2)qn1−j=µ(N)∑j=0aj(dj(q − 1)2(n1 − j) +dj(dj − 1)2)qn1−j+n1∑j=µ(N)+1aj(dj(q − 1)2(n1 − j) +dj(dj − 1)2)qn1−jthe pump journal of undergraduate research 5 (2022), 176–187 184where n1 ≥ µ(N) + 1 ≥ 1, and µ(N) will be chosen advantageously below. ThenSq =n1µ(N)∑j=0aj(dj(q − 1)2)qn1−j −µ(N)∑j=0aj(dj(q − 1)2)jqn1−j +µ(N)∑j=0aj(dj(dj − 1)2)qn1−j+O(n1qn1−µ(N))=n1n1∑j=0aj(dj(q − 1)2)qn1−j − n1n1∑j=µ(N)+1aj(dj(q − 1)2)qn1−j +O(µ(N)qn1)+O(qn1−µ(N)) +O(n1qn1−µ(N))=(q − 1)2n1n1∑j=0ajdjqn1−j +O(n1qn1−µ(N)) +O(µ(N)qn1) +O(n1qn1−µ(N))=(q − 1)2n1N +O(µ(N)qn1) +O(n1qn1−µ(N)).Consider that,N = c1qn1 + c2qn2 + c3qn3 + · · ·+ ctqntlogqN = logq(qn1(c1 + c2qn2−n1 + c3qn3−n1 + · · ·+ ctqnt−n1))logqN = n1 + logq(c1 + c2qn2−n1 + c3qn3−n1 + · · ·+ ctqnt−n1).Since1 ≤ c1 + c2qn2−n1 + c3qn3−n1 + · · ·+ ctqnt−n1 < (q − 1)∞∑i=01qi= q,we have0 ≤ logq(c1 + c2qn2−n1 + c3qn3−n1 + · · ·+ ctqnt−n1) < 1n1 ≤ logqN < n1 + 1.Thus, n1 = blogqNc =⌊logNlog q⌋.Now choose µ(N) so the two error terms are of the same magnitude:µ(N)qn1 = n1qn1−µ(N)µ(N) = n1q−µ(N)µ(N) =⌊logNlog q⌋q−µ(N).Taking logarithms on both sides, we see that, µ(N) ∼ log(logN)log qas N −→∞.the pump journal of undergraduate research 5 (2022), 176–187 185Thus,Sq =q − 12· N logNlog q+O(N log logN).Now we prove Rq ≤ N . Although each coefficient of qni in Rq looks more complicatedwith a sum, we are still able to split them in the same way as in the base 2 case. We firstwrite out the sum,Rq = c1 · c2 · qn2 + (c1 + c2)c3 · qn3 + · · ·+ (c1 + c2 + · · ·+ ct−1)ct · qnt +t∑i=1ci.Note that,c1 · c2 · qn2 + c1 · c3 · qn3 + c1 · c4 · qn4 + · · ·+ c1 · ct · qnt ≤ c1(qn1 − 1),c2 · c3 · qn3 + c2 · c4 · qn4 + · · ·+ c2 · ct · qnt ≤ c2(qn2 − 1),c3 · c4 · qn4 + · · ·+ c3 · ct · qnt ≤ c3(qn3 − 1),...ct−1 · ct · qnt ≤ ct−1(qnt−1 − 1).If we sum all the inequalities above, we haveRq ≤ c1qn1 + c2qn2 + · · ·+ ct−1qnt−1 + ct ≤ N.Note that we only have an equality if all ni are consecutive, nt = 0, and all ci = q− 1,i.e., N = qn − 1 for some n.Therefore, we have now provedAq(N) =q − 12· N logNlog q+O(N log logN).AcknowledgmentWe are thankful to Dr. Christian Ballot for his generous help and suggestions to makethis paper better.the pump journal of undergraduate research 5 (2022), 176–187 186References[1] C. Ballot, On Zeckendorf and base b digit sums, Fibonacci Quart., 51 (2013), 319–325.[2] R. Bellman and H.N. Shapiro, On a problem in additive number theory, Ann. of Math., 49 (1948),333–340.[3] L.E. Bush, An asymptotic formula for the average sum of the digits of integers, Amer. Math. Monthly,47 (1940), 154–156.[4] P. Cheo and S. Yien, A problem on the k-adic representation of positive integers, Acta Math. Sinica,5 (1955), 433–438[5] J. Pihko, An algorithm for the additive representation of positive integers, Ann. Acad. Sci. Fenn.Ser. A I Math. Dissertationes, No. 46 (1983), 54 pp.[6] L. Mirsky, A theorem on representations of integers in the scale of r, Scripta Math., 15 (1949),11–12.Cora GaddBrigham Young University - Hawaii55-220 Kulanui StreetLaie, HI, 96762E-mail: coragadd@go.byuh.eduKa Lun WongBrigham Young University - Hawaii55-220 Kulanui StreetLaie, HI, 96762E-mail: kalun.wong@byuh.eduReceived: September 15, 2022 Accepted: October 1, 2022Communicated by Steven J. Millerthe pump journal of undergraduate research 5 (2022), 176–187 187

A Generalization to Bellman and Shapiro\u27s Method on the Sum of Digital Sum Functions

https://journals.calstate.edu/pump/article/download/3485/3113

A Generalization to Bellman and Shapiro's Method on the Sum of Digital Sum Functions

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