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A051868
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16-gonal (or hexadecagonal) numbers: a(n) = n*(7*n-6).
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11
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0, 1, 16, 45, 88, 145, 216, 301, 400, 513, 640, 781, 936, 1105, 1288, 1485, 1696, 1921, 2160, 2413, 2680, 2961, 3256, 3565, 3888, 4225, 4576, 4941, 5320, 5713, 6120, 6541, 6976, 7425, 7888, 8365, 8856, 9361, 9880, 10413, 10960, 11521
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OFFSET
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0,3
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COMMENTS
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Sequence found by reading the line from 0, in the direction 0, 16, ... and the parallel line from 1, in the direction 1, 45, ..., in the square spiral whose vertices are the generalized 16-gonal numbers. - Omar E. Pol, Jul 18 2012
Let T(n) = A000217(n), the n-th triangular number. Then a(n) = T(n-1) + T(4n-3) - T(2n-4) + T(n-3). In general, let P(k,n) be the n-th k-gonal number. Then for k>1, P(T(k)+1,n) = T(n-1) + T((k-1)n-(k-2)) - T((k-3)n-2(k-3)) + T((k-4)n-3(k-4)) - ... + (-1)^(k+1)*T(n-(k-2)). - Charlie Marion, Dec 23 2019
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REFERENCES
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Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, p. 189.
E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 6.
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LINKS
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FORMULA
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a(0)=0, a(1)=1, a(2)=16; for n>2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, May 07 2011
a(14*a(n) + 92*n + 1) = a(14*a(n) + 92*n) + a(14*n+1). - Vladimir Shevelev, Jan 24 2014
a(n) = (4*n-3)^2 - (3*n-3)^2. In general, if we let P(k,n) be the n-th k-gonal number, then P(4k,n) = (k*n-k+1)^2 - ((k-1)*n-k+1)^2. In addition, {P(4k,n)} are the only polygonal number sequences each of whose terms can be written as the difference of two squares. - Charlie Marion, Feb 16 2020
Product_{n>=2} (1 - 1/a(n)) = 7/8. - Amiram Eldar, Jan 22 2021
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MATHEMATICA
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Table[n(7n-6), {n, 0, 50}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 1, 16}, 51] (* Harvey P. Dale, May 07 2011 *)
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PROG
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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