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 A194458 Total number of entries in rows 0,1,...,n of Pascal's triangle not divisible by 5. 2
 1, 3, 6, 10, 15, 17, 21, 27, 35, 45, 48, 54, 63, 75, 90, 94, 102, 114, 130, 150, 155, 165, 180, 200, 225, 227, 231, 237, 245, 255, 259, 267, 279, 295, 315, 321, 333, 351, 375, 405, 413, 429, 453, 485, 525, 535, 555, 585, 625, 675, 678, 684, 693, 705, 720, 726 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS The number of zeros in the first n rows is binomial(n+1,2) - a(n). LINKS Alois P. Heinz, Table of n, a(n) for n = 0..10000 FORMULA 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. EXAMPLE 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. MAPLE a:= proc(n) local l, m, h, j;       m:= n+1;       l:= [];       while m>0 do l:= [l[], irem (m, 5, 'm')+1] od;       h:= 0;       for j to nops(l) do h:= h*l[j] +binomial (l[j], 2) *15^(j-1) od:       h     end: seq (a(n), n=0..100); MATHEMATICA 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 *) CROSSREFS A006046(n+1) = A006046(n) + A001316(n) for p=2. A006048(n+1) = A006048(n) + A006047(n+1) for p=3. a(n+1) = a(n) + A194459(n+1) for p=5. Sequence in context: A109804 A231672 A120993 * A083266 A334130 A054731 Adjacent sequences:  A194455 A194456 A194457 * A194459 A194460 A194461 KEYWORD nonn AUTHOR Paul Weisenhorn, Aug 24 2011 EXTENSIONS Edited by Alois P. Heinz, Sep 06 2011 STATUS approved

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Last modified September 28 21:47 EDT 2020. Contains 337414 sequences. (Running on oeis4.)