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 A228196 A triangle formed like Pascal's triangle, but with n^2 on the left border and 2^n on the right border instead of 1. 30
 0, 1, 2, 4, 3, 4, 9, 7, 7, 8, 16, 16, 14, 15, 16, 25, 32, 30, 29, 31, 32, 36, 57, 62, 59, 60, 63, 64, 49, 93, 119, 121, 119, 123, 127, 128, 64, 142, 212, 240, 240, 242, 250, 255, 256, 81, 206, 354, 452, 480, 482, 492, 505, 511, 512, 100, 287, 560, 806, 932 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS The third row is (n^4-n^2+24*n+24)/12. For a closed-form formula for generalized Pascal's triangle see A228576.  - Boris Putievskiy, Sep 04 2013 LINKS Boris Putievskiy, Rows n = 1..140 of triangle, flattened Rely Pellicer, David Alvo, Modified Pascal Triangle and Pascal Surfaces p.4 FORMULA T(n,0)=2^n, n>0; T(0,k) = k^2; T(n, k) = T(n-1, k-1) + T(n-1, k) for n,k >0. Closed-form formula for general case. Let L(m) and R(m) be the left border and the right border of Pascal like triangle, respectively. We denote binomial(n,k) by C(n,k). As table read by antidiagonals T(n,k)=sum_{m1=1..n}R(m1)*C(n+k-m1-1,n-m1)+sum_{m2=1..k}L(m2)*C(n+k-m2-1,k-m2); n,k >=0. As linear sequence a(n)=sum_{m1=1..i}R(m1)*C(i+j-m1-1,i-m1)+sum_{m2=1..j}L(m2)*C(i+j-m2-1,j-m2), where  i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2); n>0. Some special cases. If  L(m)={b,b,b...} b*A000012, then the second sum takes form b*C(n+k-1,j). If  L(m) is {0,b,2b,...} b*A001477, then the second sum takes form b*C(n+k,n-1). Similarly for R(m) and the first sum. For this sequence  L(m)=m^2 and R(m)=2^m. As table read by antidiagonals T(n,k)=sum_{m1=1..n}(2^m1)*C(n+k-m1-1,n-m1)+sum_{m2=1..k}(m2^2)*C(n+k-m2-1,k-m2); n,k >=0. As linear sequence a(n)=sum_{m1=1..i}(2^m1)*C(i+j-m1-1,i-m1)+sum_{m2=1..j}(m2^2)*C(i+j-m2-1,j-m2), where  i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2). EXAMPLE The start of the sequence as triangle array read by rows: 0; 1,2; 4,3,4; 9,7,7,8; 16,16,14,15,16; 25,32,30,29,31,32; 36,57,62,59,60,63,64; PROG (Python) def funcL(n):    q = n**2    return q def funcR(n):    q = 2**n    return q for n in range (1, 9871):    t=int((math.sqrt(8*n-7) - 1)/ 2)    i=n-t*(t+1)/2-1    j=(t*t+3*t+4)/2-n-1    sum1=0    sum2=0    for m1 in range (1, i+1):       sum1=sum1+funcR(m1)*binomial(i+j-m1-1, i-m1)    for m2 in range (1, j+1):       sum2=sum2+funcL(m2)*binomial(i+j-m2-1, j-m2)    sum=sum1+sum2 CROSSREFS Cf. We denote Pascal like triangle with L(n) on the left border and R(n) on the right border by (L(n),R(n)). A007318 (1,1), A008949 (1,2^n), A029600 (2,3), A029618 (3,2), A029635 (1,2), A029653 (2,1), A037027 (Fibonacci(n),1), A051601 (n,n) n>=0, A051597 (n,n) n>0, A051666 (n^2,n^2), A071919 (1,0), A074829 (Fibonacci(n), Fibonacci(n)), A074909 (1,n), A093560 (3,1), A093561 (4,1), A093562 (5,1), A093563 (6,1), A093564 (7,1), A093565 (8,1), A093644 (9,1), A093645 (10,1), A095660 (1,3), A095666 (1,4), A096940 (1,5), A096956 (1,6), A106516 (3^n,1), A108561(1,(-1)^n), A132200 (4,4), A134636 (2n+1,2n+1), A137688 (2^n,2^n), A160760 (3^(n-1),1), A164844(1,10^n), A164847 (100^n,1), A164855 (101*100^n,1), A164866 (101^n,1), A172171 (1,9), A172185 (9,11), A172283 (-9,11), A177954 (int(n/2),1), A193820 (1,2^n), A214292 (n,-n), A227074 (4^n,4^n), A227075 (3^n,3^n), A227076 (5^n,5^n), A227550 (n!,n!), A228053 ((-1)^n,(-1)^n), A228074 (Fibonacci(n), n). Cf. A000290 (row 1), A153056 (row 2), A000079 (column 1), A000225 (column 2), A132753 (column 3), A118885 (row sums of triangle array + 1), A228576 (generalized Pascal's triangle). Sequence in context: A328486 A080001 A178938 * A261326 A136743 A011387 Adjacent sequences:  A228193 A228194 A228195 * A228197 A228198 A228199 KEYWORD nonn,tabl AUTHOR Boris Putievskiy, Aug 15 2013 EXTENSIONS Cross-references corrected and extended by Philippe Deléham, Dec 27 2013 STATUS approved

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Last modified October 22 19:53 EDT 2019. Contains 328319 sequences. (Running on oeis4.)