A214458 -id:A214458 - cp's OEIS frontend

A217971 a(n) = 2^(2n+1) (2n+1)n^(2*n).

24, 2560, 653184, 301989888, 220000000000, 231818611654656, 333360204766740480, 627189298506124754944, 1495163506861268427866112, 4404019200000000000000000000, 15705682358754099640245749284864, 66686788842514206222454073642188800, 332430457331186494783020411573611003904

Offset: 1

Vladimir Shevelev and Peter J. C. Moses, Oct 16 2012

nonn

Let S_(2*n+1)(m) denote difference between multiples of 2*n+1 in interval [0,m), m>=1, with even and odd digit sums in base 2*n. As is shown in the Shevelev and Moses link, a recursion for S_(2*n+1)(m) is connected with the periodicity of a special digit function, the smallest period of which is a(n).

Andrew Howroyd, Table of n, a(n) for n = 1..100
Vladimir Shevelev and Peter J. C. Moses, A family of digit functions with large periods, arXiv:1209.5705 [math.NT], 2012.

Cf. A214458, A218085.

Table[2^(2*n + 1)*(2*n + 1)*n^(2*n), {n, 15}] (* Wesley Ivan Hurt, Apr 28 2020 *)

A217971(n):=2^(2*n+1)*(2*n+1)*n^(2*n)$ makelist(A217971(n),n,1,10); /* Martin Ettl, Nov 15 2012 */

a(n) = {2^(2*n+1) * (2*n+1)*n^(2*n)} \\ Andrew Howroyd, Apr 28 2020

Terms a(11) and beyond from Andrew Howroyd, Apr 28 2020

A218085 Let S_5(x) denote the difference in counts of multiples of 5 in the interval [0,x), those with even digit sums in base 4 in one set, those with odd digit sums in base 4 in the other. Then a(n) = (-1)^s_4(n) (S_5(n) -10S_5(floor(n/16)) +5*S_5(floor(n/256))), where s_4(n) = A053737(n).

Original entry on oeis.org

0, -1, 1, -1, -1, 1, -2, 2, 2, -2, 2, -3, -3, 3, -3, 3, 6, -6, 6, -6, -6, 5, -5, 5, 5, -5, 4, -4, -4, 4, -4, 3, -3, 3, -3, 3, 4, -4, 4, -4, -4, 3, -3, 3, 3, -3, 2, -2, 2, -2, 2, -3, -3, 3, -3, 3, 4, -4, 4, -4, -4, 3, -3, 3, 3, -3, 2, -2, -2, 2, -2, 1, 1, -1, 1, -1, 0, 0, 0, 0, 0

Offset: 0

Vladimir Shevelev and Peter J. C. Moses, Oct 20 2012

The sequence S_5(n) starts 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, ... for n >= 0. Apart from the initial 0, these are blocks of 5 repetitions of 1, 2, 3, 4, 5, 6, 7, 6, 7, 8, 7, 6, 7, 8, 9, 10, 11, 12, 13, 14, ...
Theorem. The sequence is periodic with period 2560.
The theorem allows us to write a recursion for S_5(n), considering n modulo 2560: S_5(n) = 10*S_5(floor(n/16)) - 5*S_5(floor(n/256)) + (-1)^s_4(n)*a(n).

			a(n)=-9 for n=2411, 2412, 2414, 2491, 2492, 2494 (mod 2560);
a(n)=9 for n=2413, 2415, 2493, 2495 (mod 2560).

Peter J. C. Moses, Table of n, a(n) for n = 0..2559
Vladimir Shevelev and Peter J. C. Moses, A family of digit functions with large periods, arXiv:1209.5705 [math.NT], 2012.

Cf. A214458, A217971, A053737.

S := proc(n,j,x)
    a := 0 ;
    for r from j to x-1 by n do
        add(d,d=convert(r,base,n-1)) ;
        a := a+(-1)^% ;
    end do:
    a ;
end proc:
A218085 := proc(n)
    S(5,0,n)-10*S(5,0,floor(n/16))+5*S(5,0,floor(n/256)) ;
    %*(-1)^A053737(n) ;
end proc:
seq(A218085(n),n=0..80) ; # R. J. Mathar, Oct 31 2012

-9 <= a(n) <= 9, all 19 values are actually achieved.

cp's OEIS Frontend

A217971 a(n) = 2^(2n+1) (2n+1)n^(2*n).

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A218085 Let S_5(x) denote the difference in counts of multiples of 5 in the interval [0,x), those with even digit sums in base 4 in one set, those with odd digit sums in base 4 in the other. Then a(n) = (-1)^s_4(n) (S_5(n) -10S_5(floor(n/16)) +5*S_5(floor(n/256))), where s_4(n) = A053737(n).

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cp's OEIS Frontend

A217971 a(n) = 2^(2*n+1) * (2*n+1)*n^(2*n).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Maxima

PARI

Extensions

A218085 Let S_5(x) denote the difference in counts of multiples of 5 in the interval [0,x), those with even digit sums in base 4 in one set, those with odd digit sums in base 4 in the other. Then a(n) = (-1)^s_4(n) *(S_5(n) -10*S_5(floor(n/16)) +5*S_5(floor(n/256))), where s_4(n) = A053737(n).

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Programs

Maple

Formula

A217971 a(n) = 2^(2n+1) (2n+1)n^(2*n).

A218085 Let S_5(x) denote the difference in counts of multiples of 5 in the interval [0,x), those with even digit sums in base 4 in one set, those with odd digit sums in base 4 in the other. Then a(n) = (-1)^s_4(n) (S_5(n) -10S_5(floor(n/16)) +5*S_5(floor(n/256))), where s_4(n) = A053737(n).