A081494
Start with Pascal's triangle; form a triangle by sliding down n steps from top on both sides and including the horizontal row, deleting the inner numbers; a(n) = sum of entries on perimeter of triangle.
Original entry on oeis.org
1, 3, 7, 13, 23, 41, 75, 141, 271, 529, 1043, 2069, 4119, 8217, 16411, 32797, 65567, 131105, 262179, 524325, 1048615, 2097193, 4194347, 8388653, 16777263, 33554481, 67108915, 134217781, 268435511, 536870969, 1073741883, 2147483709
Offset: 1
The triangle pertaining to n = 4 is obtained from the solid triangle
1
1 1
1 2 1
1 3 3 1
giving
1
1 1
1 1
1 3 3 1
and the sum of all the numbers is 13, so a(4) = 13.
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restart:a:= proc(n) option remember; if n=0 then 1 else add((binomial (n,j)+2), j=0..n-1) fi end: seq (a(n), n=0..31); # Zerinvary Lajos, Mar 29 2009
A081495
Start with Pascal's triangle; form a rhombus by sliding down n steps from top on both sides then sliding down inwards to complete the rhombus and then deleting the inner numbers; a(n) = sum of entries on perimeter of rhombus.
Original entry on oeis.org
1, 5, 17, 55, 189, 681, 2519, 9451, 35765, 136153, 520695, 1998745, 7696467, 29716025, 115000947, 445962899, 1732525861, 6741529113, 26270128535, 102501265057, 400411345659, 1565841089321, 6129331763923, 24014172955545, 94163002754699, 369507926510401
Offset: 1
The rhombus pertaining to n = 4 is obtained from the solid rhombus
.....1
...1...1
.1...2...1
1..3...3...1
..4..6...4
...10..10
.....20
giving
.....1
...1...1
.1.......1
1..........1
..4......4
...10..10
.....20
and the sum of all the numbers is 55, a(4) = 55.
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B:=Binomial;; Concatenation([1], List([2..25], n-> B(2*n, n)-B(2*(n-1), n-1) +2*n -3)); # G. C. Greubel, Aug 13 2019
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C:=Catalan; [1] cat [(n+1)*C(n) -n*C(n-1) +2*n-3: n in [2..25]]; // G. C. Greubel, Aug 13 2019
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seq(coeff(series(((1-x)^3 - (1-2*x-x^3)*sqrt(1-4*x))/((1-x)^2*sqrt(1-4*x) ), x, n+1), x, n), n = 1..25); # G. C. Greubel, Aug 13 2019
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With[{C = CatalanNumber}, Table[If[n==1, 1, (n+1)*C[n] -n*C[n-1] +2*n-3], {n, 25}]] (* G. C. Greubel, Aug 13 2019 *)
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vector(25, n, b=binomial; if(n==1,1,b(2*n, n)-b(2*(n-1), n-1) +2*n -3)) \\ G. C. Greubel, Aug 13 2019
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b=binomial; [1]+[b(2*n, n)-b(2*(n-1), n-1) +2*n -3 for n in (2..25)] # G. C. Greubel, Aug 13 2019
A081496
Start with Pascal's triangle; a(n) is the sum of the numbers on the periphery of the n-th central rhombus containing exactly 4 numbers.
Original entry on oeis.org
5, 14, 46, 160, 574, 2100, 7788, 29172, 110110, 418132, 1595620, 6113744, 23505356, 90633800, 350351640, 1357278300, 5268292830, 20483876820, 79765662900, 311038321440, 1214362277700, 4746455801880, 18570960418920, 72728638093800
Offset: 1
The first three rhombuses are
...1...........2.........6
.1...1.......3...3.....10..10
...2......,....6.....,...20
and the corresponding sums are a(1) =5, a(2) =14 and a(3) =46.
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seq((9*n-4)*binomial(2*(n-1),(n-1))/n,n=1..26); # C. Ronaldo, Dec 20 2004
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{ A029635(n, k) = if( k<0 || k>n, 0, (n==0) + binomial(n, k) + binomial(n-1, k-1))}; \\ program from Michael Somos in A029635
{a(n) = sum(k=0,n,A029635(n, k)^2)} \\ Paul D. Hanna, Oct 17 2017
for(n=1,30,print1(a(n),", "))
More terms from C. Ronaldo (aga_new_ac(AT)hotmail.com), Dec 20 2004
Showing 1-3 of 3 results.