A213500
Rectangular array T(n,k): (row n) = b**c, where b(h) = h, c(h) = h + n - 1, n >= 1, h >= 1, and ** = convolution.
Original entry on oeis.org
1, 4, 2, 10, 7, 3, 20, 16, 10, 4, 35, 30, 22, 13, 5, 56, 50, 40, 28, 16, 6, 84, 77, 65, 50, 34, 19, 7, 120, 112, 98, 80, 60, 40, 22, 8, 165, 156, 140, 119, 95, 70, 46, 25, 9, 220, 210, 192, 168, 140, 110, 80, 52, 28, 10, 286, 275, 255, 228, 196, 161, 125, 90
Offset: 1
Northwest corner (the array is read by southwest falling antidiagonals):
1, 4, 10, 20, 35, 56, 84, ...
2, 7, 16, 30, 50, 77, 112, ...
3, 10, 22, 40, 65, 98, 140, ...
4, 13, 28, 50, 80, 119, 168, ...
5, 16, 34, 60, 95, 140, 196, ...
6, 19, 40, 70, 110, 161, 224, ...
T(6,1) = (1)**(6) = 6;
T(6,2) = (1,2)**(6,7) = 1*7+2*6 = 19;
T(6,3) = (1,2,3)**(6,7,8) = 1*8+2*7+3*6 = 40.
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b[n_] := n; c[n_] := n
t[n_, k_] := Sum[b[k - i] c[n + i], {i, 0, k - 1}]
TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]]
Flatten[Table[t[n - k + 1, k], {n, 12}, {k, n, 1, -1}]]
r[n_] := Table[t[n, k], {k, 1, 60}] (* A213500 *)
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t(n,k) = sum(i=0, k - 1, (k - i) * (n + i));
tabl(nn) = {for(n=1, nn, for(k=1, n, print1(t(k,n - k + 1),", ");); print(););};
tabl(12) \\ Indranil Ghosh, Mar 26 2017
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def t(n, k): return sum((k - i) * (n + i) for i in range(k))
for n in range(1, 13):
print([t(k, n - k + 1) for k in range(1, n + 1)]) # Indranil Ghosh, Mar 26 2017
A100157
Structured rhombic dodecahedral numbers (vertex structure 9).
Original entry on oeis.org
1, 14, 55, 140, 285, 506, 819, 1240, 1785, 2470, 3311, 4324, 5525, 6930, 8555, 10416, 12529, 14910, 17575, 20540, 23821, 27434, 31395, 35720, 40425, 45526, 51039, 56980, 63365, 70210, 77531, 85344, 93665, 102510, 111895, 121836, 132349, 143450, 155155, 167480
Offset: 1
James A. Record (james.record(AT)gmail.com), Nov 07 2004
For n=4, sum( (4+i)^2, i=-3..3 ) = (4-3)^2+(4-2)^2+(4-1)^2+(4-0)^2+(4+1)^2+(4+2)^2+(4+3)^2 = 140 = a(4). - _Bruno Berselli_, Jul 24 2014
- Jolley, Summation of Series, Dover (1961).
- Vincenzo Librandi, Table of n, a(n) for n = 1..5000
- Milan Janjic, Two Enumerative Functions.
- A. Schuetz and G. Whieldon, Polygonal Dissections and Reversions of Series, arXiv:1401.7194 [math.CO], 2014.
- StackExchange, What is a Structured Polyhedron?
- Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Cf.
A005915 = alternate vertex;
A100145 for more on structured polyhedral numbers.
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[(1/6)*(16*n^3-12*n^2+2*n): n in [1..40]]; // Vincenzo Librandi, Jul 19 2011
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with(combstruct):ZL:=[st, {st=Prod(left, right), left=Set(U, card=r), right=Set(U, card=r), U=Sequence(Z, card>=1)}, unlabeled]: subs(r=1, stack): seq(count(subs(r=2, ZL), size=m*4), m=1..32) ; # Zerinvary Lajos, Jan 02 2008
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a(n)=(16*n^3-12*n^2+2*n)/6 \\ Charles R Greathouse IV, Sep 24 2015
A071238
a(n) = n*(n+1)*(2*n^2+1)/6.
Original entry on oeis.org
0, 1, 9, 38, 110, 255, 511, 924, 1548, 2445, 3685, 5346, 7514, 10283, 13755, 18040, 23256, 29529, 36993, 45790, 56070, 67991, 81719, 97428, 115300, 135525, 158301, 183834, 212338, 244035, 279155, 317936, 360624, 407473, 458745, 514710, 575646, 641839
Offset: 0
- T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.
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[n*(n+1)*(2*n^2+1)/6: n in [0..40]]; // Vincenzo Librandi, Jun 14 2011
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A071238:=n->n*(n+1)*(2*n^2+1)/6: seq(A071238(n), n=0..60); # Wesley Ivan Hurt, Sep 24 2016
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Table[n (n + 1) (2 n^2 + 1)/6, {n, 0, 37}] (* or *)
CoefficientList[Series[x (1 + x) (1 + 3 x)/(1 - x)^5, {x, 0, 37}], x] (* Michael De Vlieger, Sep 14 2016 *)
LinearRecurrence[{5,-10,10,-5,1},{0,1,9,38,110},40] (* Harvey P. Dale, Oct 02 2021 *)
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a(n)=n*(n+1)*(2*n^2+1)/6; \\ Joerg Arndt, Sep 04 2013
A143941
The Wiener index of a chain of n triangles (i.e., joined like VVV..VV; here V is a triangle!).
Original entry on oeis.org
3, 14, 37, 76, 135, 218, 329, 472, 651, 870, 1133, 1444, 1807, 2226, 2705, 3248, 3859, 4542, 5301, 6140, 7063, 8074, 9177, 10376, 11675, 13078, 14589, 16212, 17951, 19810, 21793, 23904, 26147, 28526, 31045, 33708, 36519, 39482, 42601, 45880, 49323, 52934
Offset: 1
a(2)=14 because in the graph VV (V is a triangle!) we have 6 distances equal to 1 and 4 distances equal to 2.
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[n*(1+6*n+2*n^2)/3 : n in [1..40]]; // Wesley Ivan Hurt, Apr 08 2015
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seq((1/3)*n*(1+6*n+2*n^2), n=1..43);
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CoefficientList[Series[(3+2*x-x^2)/(1-x)^4,{x,0,40}],x] (* Vincenzo Librandi, Jul 03 2012 *)
LinearRecurrence[{4,-6,4,-1},{3,14,37,76},50] (* Harvey P. Dale, Sep 06 2023 *)
Showing 1-4 of 4 results.
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