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 A121358 Least prime factor of pyramidal number A000292(n), a(1) = 1. 1
 1, 2, 2, 2, 5, 2, 2, 2, 3, 2, 2, 2, 5, 2, 2, 2, 3, 2, 2, 2, 7, 2, 2, 2, 3, 2, 2, 2, 5, 2, 2, 2, 5, 2, 2, 2, 13, 2, 2, 2, 7, 2, 2, 2, 3, 2, 2, 2, 5, 2, 2, 2, 3, 2, 2, 2, 19, 2, 2, 2, 3, 2, 2, 2, 5, 2, 2, 2, 5, 2, 2, 2, 5, 2, 2, 2, 7, 2, 2, 2, 3, 2, 2, 2, 5, 2, 2, 2, 3, 2, 2, 2, 5, 2, 2, 2, 3, 2, 2, 2, 17, 2, 2, 2, 5 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS For n={3,4,5}+4*k, k=0,1,..., a(n)=2. If we omit these terms we get sequence a(2+4*n) = 5, 3, 5, 3, 7, 3, 5, 5, 13, 7, 3, 5, 3, 19, 3, 5, 5, 5, 7, 3, 5, 3, 5, 3, 17, 5, 5, 5, 3, 11, 3, 5, 3, 23, 11, 5, 5, 3, 53, 3, 5, 3, 5, 59, 7, 5, 3, 5, 3, 7, 3, 5, 5, 7, 13, 3, 5, 3, 7, 3,  5, 5, 5, 7, 3, 5, 3, 5, 3, 47, 5, 5, 5, 3; least prime factor of (1 + 4*n)*(2 + 4*n)*(3 + 4*n)/6, n=1,2,... Cf. A000292 Tetrahedral (or pyramidal) numbers: C(n+2,3) = n(n+1)(n+2)/6. LINKS Antti Karttunen, Table of n, a(n) for n = 1..65537 FORMULA a(n) = lpf(n(n+1)(n+2)/6), for n >= 2, with a(1) = 1. a(n) = A020639(A000292(n)). - Antti Karttunen, Jul 22 2018 MATHEMATICA FactorInteger[#][[1, 1]]&/@Binomial[Range[2, 110]+2, 3] (* Harvey P. Dale, Dec 07 2016 *) PROG (PARI) A000292(n) = (n*(n+1)*(n+2)/6); A020639(n) = if(1==n, n, factor(n)[1, 1]); A121358(n) = A020639(A000292(n)); \\ Antti Karttunen, Jul 22 2018 CROSSREFS Cf. A000292, A020639. Sequence in context: A088885 A232736 A275887 * A271321 A112659 A115281 Adjacent sequences:  A121355 A121356 A121357 * A121359 A121360 A121361 KEYWORD nonn AUTHOR Zak Seidov, Sep 06 2006 EXTENSIONS Term a(1) = 1 prepended and offset corrected by Antti Karttunen, Jul 22 2018 STATUS approved

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Last modified June 17 19:10 EDT 2019. Contains 324198 sequences. (Running on oeis4.)