|
%I #39 Nov 02 2022 11:53:39
%S 1,3,6,10,15,17,21,27,35,45,48,54,63,75,90,94,102,114,130,150,155,165,
%T 180,200,225,227,231,237,245,255,259,267,279,295,315,321,333,351,375,
%U 405,413,429,453,485,525,535,555,585,625,675,678,684,693,705,720,726
%N Total number of entries in rows 0,1,...,n of Pascal's triangle not divisible by 5.
%C The number of zeros in the first n rows is binomial(n+1,2) - a(n).
%H Alois P. Heinz, <a href="/A194458/b194458.txt">Table of n, a(n) for n = 0..10000</a>
%H Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, <a href="https://arxiv.org/abs/2210.10968">Identities and periodic oscillations of divide-and-conquer recurrences splitting at half</a>, arXiv:2210.10968 [cs.DS], 2022, p. 53.
%F a(n) = ((C(d0+1,2)*15^0*(d1+1) + C(d1+1,2)*15^1)*(d1+1) + C(d1+1,2)*15^1)*(d2+1) + C(d2+1,2)*15^2 ..., where d_k...d_1d_0 is the base 5 expansion of n+1 and 15 = binomial(5+1,2). The formula generalizes to other prime bases p.
%e n = 38: n+1 = 39 = 124_5, thus a(38) = (C(5,2)*15^0*3 + C(3,2)*15^1)*2 + C(2,2)*15^2 = (10*1*3 + 3*15)*2 + 1*225 = 375.
%p a:= proc(n) local l, m, h, j;
%p m:= n+1;
%p l:= [];
%p while m>0 do l:= [l[], irem (m, 5, 'm')+1] od;
%p h:= 0;
%p for j to nops(l) do h:= h*l[j] +binomial (l[j], 2) *15^(j-1) od:
%p h
%p end:
%p seq (a(n), n=0..100);
%t a[n_] := Module[{l, m, r, h, j}, m = n+1; l = {}; While[m>0, l = Append[l, {m, r} = QuotientRemainder[m, 5]; r+1]]; h = 0; For[j = 1, j <= Length[l], j++, h = h*l[[j]] + Binomial [l[[j]], 2] *15^(j-1)]; h]; Table [a[n], {n, 0, 100}] (* _Jean-François Alcover_, Feb 26 2017, translated from Maple *)
%Y A006046(n+1) = A006046(n) + A001316(n) for p=2.
%Y A006048(n+1) = A006048(n) + A006047(n+1) for p=3.
%Y a(n+1) = a(n) + A194459(n+1) for p=5.
%K nonn
%O 0,2
%A _Paul Weisenhorn_, Aug 24 2011
%E Edited by _Alois P. Heinz_, Sep 06 2011
|
