A092093 Back and Forth Summant S(n, 5): a(n) = sum{i = 0..floor(2n/5)} n-5i.
1, 2, 1, 3, 0, 3, 6, 2, 6, 0, 5, 10, 3, 9, 0, 7, 14, 4, 12, 0, 9, 18, 5, 15, 0, 11, 22, 6, 18, 0, 13, 26, 7, 21, 0, 15, 30, 8, 24, 0, 17, 34, 9, 27, 0, 19, 38, 10, 30, 0, 21, 42, 11, 33, 0, 23, 46, 12, 36, 0, 25, 50, 13, 39, 0, 27, 54, 14, 42, 0, 29, 58, 15, 45, 0, 31, 62, 16, 48, 0, 33
Offset: 1
References
- J. Dezert, editor, Smarandacheials, Mathematics Magazine, Aurora, Canada, No. 4/2004.
- F. Smarandache, Back and Forth Factorials, Arizona State Univ., Special Collections, 1972.
- F. Smarandache, Back and Forth Summants, Arizona State Univ., Special Collections, 1972.
Links
- J. Dezert, Smaran dacheials
- J. Dezert, S marandacheials, "Mathematics Magazine", Canada
- F. Smarandache, Summants
- Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,2,0,0,0,0,-1).
Crossrefs
Programs
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PARI
S(n, k=5) = local(s, x); s = n; x = n - k; while (x >= -n, s = s + x; x = x - k); s;
Formula
a(5n) = 0; a(5n+1) = 2n+1; a(5n+2) = 4n+2; a(5n+3) = n+1; a(5n+4) = 3n+3.
G.f.: x*(2*x^6+x^5+3*x^3+x^2+2*x+1) / ((x-1)^2*(x^4+x^3+x^2+x+1)^2). - Colin Barker, Jul 28 2013
Extensions
Edited and extended by David Wasserman, Dec 19 2005