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 A152545 Padovan-Fibonacci triangle, read by rows, where the first column equals the Padovan spiral numbers (A134816), while the row sums equal the Fibonacci numbers (A000045). 3
 1, 1, 1, 1, 2, 1, 2, 2, 1, 3, 2, 2, 1, 4, 3, 3, 2, 1, 5, 4, 4, 3, 3, 1, 1, 7, 5, 5, 5, 4, 3, 3, 1, 1, 9, 7, 7, 7, 5, 5, 5, 4, 3, 1, 1, 1, 12, 9, 9, 9, 8, 7, 7, 7, 5, 4, 4, 4, 1, 1, 1, 1, 16, 12, 12, 12, 12, 9, 9, 9, 8, 8, 8, 7, 5, 4, 4, 4, 1, 1, 1, 1, 1, 21, 16, 16, 16, 16, 13, 12, 12, 12, 12, 12, 11, 9, 8 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,5 COMMENTS The number of terms in each row equal the Padovan spiral numbers (A134816, with offset). LINKS Paul D. Hanna, Table of n, a(n) for n = 0..261 FORMULA G.f. for row n: Sum_{k=0..A000931(n+5)-1} (x^{T(n-1,k)+T(n-2,k)} - 1)/(x-1) = Sum_{k=0..A000931(n+6)-1} T(n,k)*x^k for n>1 with T(0,0)=T(1,0)=1, where A000931 is the Padovan sequence. EXAMPLE Triangle begins: [1], [1], [1,1], [2,1], [2,2,1], [3,2,2,1], [4,3,3,2,1], [5,4,4,3,3,1,1], [7,5,5,5,4,3,3,1,1], [9,7,7,7,5,5,5,4,3,1,1,1], [12,9,9,9,8,7,7,7,5,4,4,4,1,1,1,1], [16,12,12,12,12,9,9,9,8,8,8,7,5,4,4,4,1,1,1,1,1], [21,16,16,16,16,13,12,12,12,12,12,11,9,8,8,8,5,5,5,5,4,1,1,1,1,1,1,1], [28,21,21,21,21,20,16,16,16,16,16,16,13,13,12,12,12,12,11,11,8,6,5,5,5,5,5,5,1,1,1,1,1,1,1,1,1], [37,28,28,28,28,28,22,21,21,21,21,21,20,20,20,20,18,16,16,16,14,13,12,12,12,12,12,11,6,6,6,6,6,5,5,5,5,1,1,1,1,1,1,1,1,1,1,1,1], [49,37,37,37,37,37,33,28,28,28,28,28,28,28,28,28,27,22,22,21,21,21,20,20,20,20,20,18,17,17,16,16,14,12,12,12,12,7,6,6,6,6,6,6,6,6,6,6,5,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1], ... ILLUSTRATION OF RECURRENCE. Start out with row 0 and row 1 consisting of a single '1'. To obtain any given row of this irregular triangle, first sum the prior two rows term-by-term; for rows 7 and 8 we get: [5,4,4,3,3,1,1] + [7,5,5,5,4,3,3,1,1] = [12,9,9,8,7,4,4,1,1]. Place markers in an array so that the number of contiguous markers in each row correspond to the term-by-term sums like so: -------------------------- 12:o o o o o o o o o o o o 9: o o o o o o o o o - - - 9: o o o o o o o o o - - - 8: o o o o o o o o - - - - 7: o o o o o o o - - - - - 4: o o o o - - - - - - - - 4: o o o o - - - - - - - - 1: o - - - - - - - - - - - 1: o - - - - - - - - - - - -------------------------- Then count the markers by columns to obtain the desired row; here, the number of markers in each column yields row 9: [9,7,7,7,5,5,5,4,3,1,1,1]. Continuing in this way generates all the rows of this triangle. PROG (PARI) {T(n, k)=local(G000931=(1-x^2)/(1-x^2-x^3+x*O(x^(n+6)))); if(n<0, 0, if(n<2&k==0, 1, polcoeff(sum(j=0, polcoeff(G000931, n+5)-1, (x^(T(n-1, j)+T(n-2, j)) - 1)/(x-1)), k) ))}; /* To print, use Padovan g.f. to get the number of terms in row n: */ for(n=0, 10, for(k=0, polcoeff((1-x^2)/(1-x^2-x^3+x*O(x^(n+6))), n+6)-1, print1(T(n, k), ", ")); print("")) CROSSREFS Cf. A134816, A000045, A000931; A152546 (row squared sums). Sequence in context: A205451 A208244 A160696 * A109967 A000119 A097368 Adjacent sequences:  A152542 A152543 A152544 * A152546 A152547 A152548 KEYWORD nonn,tabf AUTHOR Paul D. Hanna, Dec 13 2008 STATUS approved

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Last modified June 20 05:09 EDT 2013. Contains 226418 sequences.