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A167280 Period length 12: 0,0,1,2,4,7,4,8,7,4,8,5 (and repeat). 0
0, 0, 1, 2, 4, 7, 4, 8, 7, 4, 8, 5, 0, 0, 1, 2, 4, 7, 4, 8, 7, 4, 8, 5, 0, 0, 1, 2, 4, 7, 4, 8, 7, 4, 8, 5, 0, 0, 1, 2, 4, 7, 4, 8, 7, 4, 8, 5, 0, 0, 1, 2, 4, 7, 4, 8, 7, 4, 8, 5, 0, 0, 1, 2, 4, 7, 4, 8, 7, 4, 8, 5, 0, 0, 1, 2, 4, 7, 4, 8, 7, 4, 8, 5, 0, 0, 1, 2, 4, 7, 4, 8, 7, 4, 8, 5, 0, 0, 1, 2, 4, 7, 4, 8, 7 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,4

COMMENTS

The sum of the terms in the period is 50, so the partial sums of the sequence are also 12-periodic if reduced modulo 50 or modulo 10.

The weighted partial sums b(n) = sum_{i=0..n} a(i)*2^i obey b(n) = b(n+12) (mod 10).

Third column is A000689. (Which table or array is this referring to? R. J. Mathar, Nov 01 2009)

The set of digits in the period is the same as in A141425.

A derived sequence with terms a(n)+a(n+6) has period length 6: 4, 8, 8, 6, 12, 12 (repeat).

LINKS

Table of n, a(n) for n=0..104.

FORMULA

a(n) = A113405(n+1) mod 10.

G.f.: x^2*(1+2*x+4*x^2+7*x^3+4*x^4+8*x^5+7*x^6+4*x^7+8*x^8+5*x^9)/( (1-x)*(1+x+x^2)*(1+x)*(1-x+x^2)*(1+x^2)*(x^4-x^2+1)) [R. J. Mathar, Nov 03 2009]

a(n)=(1/396)*{190*(n mod 12)+124*[(n+1) mod 12]-107*[(n+2) mod 12]+124*[(n+3) mod 12]+58*[(n+4) mod 12]-107*[(n+5) mod 12]+124*[(n+6) mod 12]-74*[(n+7) mod 12]-41*[(n+8) mod 12]-8*[(n+9) mod 12]-8*[(n+10) mod 12]+25*[(n+11) mod 12]}, with n>=0 [From Paolo P. Lava, Nov 12 2009]

CROSSREFS

Sequence in context: A074958 A222407 A230724 * A236629 A019966 A021408

Adjacent sequences:  A167277 A167278 A167279 * A167281 A167282 A167283

KEYWORD

nonn,easy

AUTHOR

Paul Curtz, Nov 01 2009

EXTENSIONS

Edited by R. J. Mathar, Nov 05 2009

STATUS

approved

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Last modified February 20 07:28 EST 2020. Contains 332067 sequences. (Running on oeis4.)