A154105
a(n) = 12*n^2 + 18*n + 7.
Original entry on oeis.org
7, 37, 91, 169, 271, 397, 547, 721, 919, 1141, 1387, 1657, 1951, 2269, 2611, 2977, 3367, 3781, 4219, 4681, 5167, 5677, 6211, 6769, 7351, 7957, 8587, 9241, 9919, 10621, 11347, 12097, 12871, 13669, 14491, 15337, 16207, 17101, 18019, 18961, 19927, 20917, 21931
Offset: 0
a(2) = 12*2^2 + 18*2 + 7 = 91 = 6*14 + 7 = 6*A014106(2) + 7.
a(3) = a(2) + 24*3 + 6 = 91 + 72 + 6 = 169.
a(-4) = 12*4^2 - 18*4 + 7 = 127 = 2*64 - 1 = 2*A085473(3) - 1.
-
[ 12*n^2+18*n+7: n in [0..40] ];
-
Table[12*n^2 + 18*n + 7, {n, 0, 42}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2012 *)
LinearRecurrence[{3,-3,1}, {7,37,91}, 25] (* G. C. Greubel, Sep 02 2016 *)
-
a(n)=12*n^2+18*n+7 \\ Charles R Greathouse IV, Sep 02 2016
A213772
Principal diagonal of the convolution array A213771.
Original entry on oeis.org
1, 11, 42, 106, 215, 381, 616, 932, 1341, 1855, 2486, 3246, 4147, 5201, 6420, 7816, 9401, 11187, 13186, 15410, 17871, 20581, 23552, 26796, 30325, 34151, 38286, 42742, 47531, 52665, 58156, 64016, 70257, 76891, 83930, 91386, 99271, 107597, 116376, 125620, 135341
Offset: 1
- Zhu Shijie, Jade Mirror of the Four Unknowns (Siyuan yujian), Book III Guo Duo Die Gang (Piles of Fruit), Problem number 7, (1303).
- Clark Kimberling, Table of n, a(n) for n = 1..1000
- Zhu Shijie, Jade Mirror of the Four Unknowns 2, Translation by Library of Chinese classics, original from 1303.
- Wikipedia, Jade Mirror of the Four Unknowns.
- Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
-
(See A213771.)
LinearRecurrence[{4,-6,4,-1},{1,11,42,106},70] (* Harvey P. Dale, Mar 29 2025 *)
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a(n) = (4*n^3-3*n^2+n)/2; \\ Altug Alkan, Dec 16 2017
A085474
C(2*n+4,4)-C(2*n,4).
Original entry on oeis.org
1, 15, 69, 195, 425, 791, 1325, 2059, 3025, 4255, 5781, 7635, 9849, 12455, 15485, 18971, 22945, 27439, 32485, 38115, 44361, 51255, 58829, 67115, 76145, 85951, 96565, 108019, 120345, 133575, 147741, 162875, 179009, 196175, 214405, 233731
Offset: 0
-
[(16*n^3+12*n^2+14*n+3)/3: n in [0..50]]; // Vincenzo Librandi, Sep 22 2011
-
A085474:=n->(16*n^3+12*n^2+14*n+3)/3; seq(A085474(n), n=0..100); # Wesley Ivan Hurt, Nov 12 2013
-
Table[(16n^3+12n^2+14n+3)/3, {n,0,50}] (* Wesley Ivan Hurt, Nov 12 2013 *)
LinearRecurrence[{4,-6,4,-1},{1,15,69,195},40] (* Harvey P. Dale, Nov 10 2017 *)
A085475
Square array of binomial related numbers, read by antidiagonals.
Original entry on oeis.org
0, 1, 0, 1, 1, 0, 1, 5, 1, 0, 1, 10, 9, 1, 0, 1, 15, 31, 13, 1, 0, 1, 21, 69, 64, 17, 1, 0, 1, 28, 126, 195, 109, 21, 1, 0, 1, 36, 210, 456, 425, 166, 25, 1, 0, 1, 45, 330, 923, 1231, 791, 235, 29, 1, 0, 1, 55, 495, 1716, 2975, 2751, 1325, 316, 33, 1, 0, 1, 66, 715, 3003, 6427
Offset: 0
Rows begin
0 0 0 0 0 ...
1 1 1 1 1 ...
1 5 9 13 17 ...
1 10 31 64 109 ...
1 15 69 195 425 ...
A303273
Array T(n,k) = binomial(n, 2) + k*n + 1 read by antidiagonals.
Original entry on oeis.org
1, 1, 1, 1, 2, 2, 1, 3, 4, 4, 1, 4, 6, 7, 7, 1, 5, 8, 10, 11, 11, 1, 6, 10, 13, 15, 16, 16, 1, 7, 12, 16, 19, 21, 22, 22, 1, 8, 14, 19, 23, 26, 28, 29, 29, 1, 9, 16, 22, 27, 31, 34, 36, 37, 37, 1, 10, 18, 25, 31, 36, 40, 43, 45, 46, 46, 1, 11, 20, 28, 35, 41
Offset: 0
The array T(n,k) begins
1 1 1 1 1 1 1 1 1 1 1 1 1 ... A000012
1 2 3 4 5 6 7 8 9 10 11 12 13 ... A000027
2 4 6 8 10 12 14 16 18 20 22 24 26 ... A005843
4 7 10 13 16 19 22 25 28 31 34 37 40 ... A016777
7 11 15 19 23 27 31 35 39 43 47 51 55 ... A004767
11 16 21 26 31 36 41 46 51 56 61 66 71 ... A016861
16 22 28 34 40 46 52 58 64 70 76 82 88 ... A016957
22 29 36 43 50 57 64 71 78 85 92 99 106 ... A016993
29 37 45 53 61 69 77 85 93 101 109 117 125 ... A004770
37 46 55 64 73 82 91 100 109 118 127 136 145 ... A017173
46 56 66 76 86 96 106 116 126 136 146 156 166 ... A017341
56 67 78 89 100 111 122 133 144 155 166 177 188 ... A017401
67 79 91 103 115 127 139 151 163 175 187 199 211 ... A017605
79 92 105 118 131 144 157 170 183 196 209 222 235 ... A190991
...
The inverse binomial transforms of the columns are
1 1 1 1 1 1 1 1 1 1 1 1 1 ...
0 1 2 3 4 5 6 7 8 9 10 11 12 ...
1 1 1 1 1 1 1 1 1 1 1 1 1 ...
0 0 0 0 0 0 0 0 0 0 0 0 0 ...
0 0 0 0 0 0 0 0 0 0 0 0 0 ...
0 0 0 0 0 0 0 0 0 0 0 0 0 ...
...
T(k,n-k) = A087401(n,k) + 1 as triangle
1
1 1
1 2 2
1 3 4 4
1 4 6 7 7
1 5 8 10 11 11
1 6 10 13 15 16 16
1 7 12 16 19 21 22 22
1 8 14 19 23 26 28 29 29
1 9 16 22 27 31 34 36 37 37
1 10 18 25 31 36 40 43 45 46 46
...
- R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics: A Foundation for Computer Science, Addison-Wesley, 1994.
-
T := (n, k) -> binomial(n, 2) + k*n + 1;
for n from 0 to 20 do seq(T(n, k), k = 0 .. 20) od;
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Table[With[{n = m - k}, Binomial[n, 2] + k n + 1], {m, 0, 11}, {k, m, 0, -1}] // Flatten (* Michael De Vlieger, Apr 21 2018 *)
-
T(n, k) := binomial(n, 2)+ k*n + 1$
for n:0 thru 20 do
print(makelist(T(n, k), k, 0, 20));
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T(n,k) = binomial(n, 2) + k*n + 1;
tabl(nn) = for (n=0, nn, for (k=0, nn, print1(T(n, k), ", ")); print); \\ Michel Marcus, May 17 2018
A347533
Array A(n,k) where A(n,0) = n and A(n,k) = (k*n + 1)^2 - A(n,k-1), n > 0, read by ascending antidiagonals.
Original entry on oeis.org
1, 2, 3, 3, 7, 6, 4, 13, 18, 10, 5, 21, 36, 31, 15, 6, 31, 60, 64, 50, 21, 7, 43, 90, 109, 105, 71, 28, 8, 57, 126, 166, 180, 151, 98, 36, 9, 73, 168, 235, 275, 261, 210, 127, 45, 10, 91, 216, 316, 390, 401, 364, 274, 162, 55, 11, 111, 270, 409, 525, 571, 560, 477, 351, 199, 66
Offset: 1
Array, A(n, k), begins:
1 3 6 10 15 21 28 36 45 ... A000217;
2 7 18 31 50 71 98 127 162 ... A195605;
3 13 36 64 105 151 210 274 351 ...
4 21 60 109 180 261 364 477 612 ...
5 31 90 166 275 401 560 736 945 ...
6 43 126 235 390 571 798 1051 1350 ...
7 57 168 316 525 771 1078 1422 1827 ...
8 73 216 409 680 1001 1400 1849 2376 ...
9 91 270 514 855 1261 1764 2332 2997 ...
Antidiagonals, T(n, k), begin as:
1;
2, 3;
3, 7, 6;
4, 13, 18, 10;
5, 21, 36, 31, 15;
6, 31, 60, 64, 50, 21;
7, 43, 90, 109, 105, 71, 28;
8, 57, 126, 166, 180, 151, 98, 36;
9, 73, 168, 235, 275, 261, 210, 127, 45;
10, 91, 216, 316, 390, 401, 364, 274, 162, 55;
Family of sequences (k*n + 1)^2:
A016754 (k=2),
A016778 (k=3),
A016814 (k=4),
A016862 (k=5),
A016922 (k=6),
A016994 (k=7),
A017078 (k=8),
A017174 (k=9),
A017282 (k=10),
A017402 (k=11),
A017534 (k=12),
A134934 (k=14).
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A347533:= func< n,k | (1/2)*((k*(n-k)+1)*((k+1)*(n-k)+1) +(-1)^k*(n-k- 1)) >;
[A347533(n,k): k in [0..n-1], n in [1..13]]; // G. C. Greubel, Dec 25 2022
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A[n_, 0]:= n; A[n_, k_]:= (k*n+1)^2 -A[n,k-1]; Table[Function[n, A[n, k]][m-k+1], {m,0,10}, {k,0,m}]//Flatten (* Michael De Vlieger, Oct 27 2021 *)
-
def A347533(n,k): return (1/2)*((k*(n-k)+1)*((k+1)*(n-k)+1) +(-1)^k*(n-k- 1))
flatten([[A347533(n,k) for k in range(n)] for n in range(1,14)]) # G. C. Greubel, Dec 25 2022
A385108
Triangle a(n,k) read by antidiagonals: a(n,k) is the number of dots in the k-augmented centered triangle of order n, k>=0, n>=1.
Original entry on oeis.org
1, 1, 4, 1, 10, 10, 1, 31, 31, 19, 1, 109, 109, 64, 31, 1, 409, 409, 235, 109, 46, 1, 1585, 1585, 901, 409, 166, 64, 1, 6241, 6241, 3529, 1585, 631, 235, 85, 1, 24769, 24769, 13969, 6241, 2461, 901, 316, 109, 1, 98689, 98689, 55585, 24769, 9721, 3529, 1219, 409, 136
Offset: 1
For n=1, any k:
a(1,k) = 4^k*(3*1^2-3*1+2)/2 - 3*1*Sum_{j=1..k} 4^(k-j)*2^(j-1) + 3*Sum_{j=1..k} 4^{k-j}*(2^{j-1}-1)
= 4^k*1 - 3*Sum_{j=1..k} 4^(k-j)*2^(j-1) + 3*Sum_{j=1..k} 4^(k-j)*2^(j-1) - 3*Sum_{j=1..k} 4^(k-j)
= 4^k-3*Sum_{j=1..k} 4^(k-j)
= 4^k-3*(4^k-1)/3
= 4^k-4^k+1
= 1.
For n=2,k=0:
a(2,0) = 4^0*(3*2^2-3*2+2)/2
= 8/2
= 4.
For n=3, k=2:
a(3,2) = 4^2*(3*3^2-3*3+2)/2 - 3*3*Sum_{j=1..2} 4^(2-j)*2^(j-1) + 3*Sum_{j=1..22} 4^(2-j)*(2^(j-1)-1)
= 16*20/2-9*6+3*1
= 109.
Square array begins:
1, 1, 1, 1, 1, 1, 1, 1, ...
4, 10, 31, 109, 409, 1585, 6241, 24769, ...
10, 31, 109, 409, 1585, 6241, 24769, 98689, ...
19, 64, 235, 901, 3529, 13969, 55585, 221761, ...
31, 109, 409, 1585, 6241, 24769, 98689, 393985, ...
46, 166, 631, 2461, 9721, 38641, 154081, 615361, ...
64, 235, 901, 3529, 13969, 55585, 221761, 885889, ...
85, 316, 1219, 4789, 18985, 75601, 301729, 1205569, ...
-
a := proc(n, k)
local S1, S2, j;
S1 := add(4^(k-j)*2^(j-1), j=1..k);
S2 := add(4^(k-j)*(2^(j-1)-1), j=1..k);
return 4^k*(3*n^2 - 3*n + 2)/2 - 3*n*S1 + 3*S2;
end proc:
seq(seq(a(n,d-n), n=1..d), d=1..10);
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a[n_, k_] := Module[{S1, S2},
S1 = Sum[4^(k - j) * 2^(j - 1), {j, 1, k}];
S2 = Sum[4^(k - j) * (2^(j - 1) - 1), {j, 1, k}];
4^k * (3 n^2 - 3 n + 2)/2 - 3 n * S1 + 3 * S2
];
Table[a[n, k], {n, 1, 5}, {k, 0, 5}]
TableForm[Table[a[n, k], {n, 1, 5}, {k, 0, 5}]]
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def a(n, k):
return 4**k*(3*n**2-3*n+2)//2-3*n*sum(4**(k-j)*2**(j-1) for j in range(1,k+1))+ 3*sum(4**(k-j)*(2**(j-1)-1) for j in range(1,k+1))
for n in range(1, 10):
row = [a(n, k)
for k in range(0, 10)]
print(row)
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# For the antidiagonal terms.
def a(n,k):
term1 = 4**k * (3*n**2 - 3*n + 2) // 2
if k == 0:
return term1
return term1 - 3*n*sum([4**(k-j) * 2**(j-1) for j in range(1, k+1)]) + 3*sum([4**(k-j) * (2**(j-1) - 1) for j in range(1, k+1)])
def antidiagonal_sequence(num_terms):
terms = []
diag = 0
while len(terms) < num_terms:
for n in range(1, diag+2):
k = diag - (n-1)
if k < 0:
continue
terms.append(a(n, k))
if len(terms) == num_terms:
break
diag += 1
return terms
N = 55
seq = antidiagonal_sequence(N)
print(', '.join(str(x) for x in seq))
A115519
n*(1+3*n+6*n^2)/2.
Original entry on oeis.org
0, 5, 31, 96, 218, 415, 705, 1106, 1636, 2313, 3155, 4180, 5406, 6851, 8533, 10470, 12680, 15181, 17991, 21128, 24610, 28455, 32681, 37306, 42348, 47825, 53755, 60156, 67046, 74443, 82365, 90830, 99856, 109461, 119663, 130480, 141930, 154031, 166801, 180258
Offset: 0
A338369
Triangle read by rows: T(n,k) = (Sum_{i=0..n-k}(1+k*i)^3)/(Sum_{i=0..n-k} (1+k*i)) for 0 <= k <= n.
Original entry on oeis.org
1, 1, 1, 1, 3, 1, 1, 6, 7, 1, 1, 10, 17, 13, 1, 1, 15, 31, 34, 21, 1, 1, 21, 49, 64, 57, 31, 1, 1, 28, 71, 103, 109, 86, 43, 1, 1, 36, 97, 151, 177, 166, 121, 57, 1, 1, 45, 127, 208, 261, 271, 235, 162, 73, 1, 1, 55, 161, 274, 361, 401, 385, 316, 209, 91, 1, 1, 66, 199, 349, 477, 556, 571, 519, 409, 262, 111, 1
Offset: 0
The triangle T(n,k) for 0 <= k <= n starts:
n \k : 0 1 2 3 4 5 6 7 8 9 10 11 12
====================================================================
0 : 1
1 : 1 1
2 : 1 3 1
3 : 1 6 7 1
4 : 1 10 17 13 1
5 : 1 15 31 34 21 1
6 : 1 21 49 64 57 31 1
7 : 1 28 71 103 109 86 43 1
8 : 1 36 97 151 177 166 121 57 1
9 : 1 45 127 208 261 271 235 162 73 1
10 : 1 55 161 274 361 401 385 316 209 91 1
11 : 1 66 199 349 477 556 571 519 409 262 111 1
12 : 1 78 241 433 609 736 793 771 673 514 321 133 1
etc.
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T[n_, k_] := Sum[(1 + k*i)^3, {i, 0, n - k}]/Sum[1 + k*i, {i, 0, n - k}]; Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Amiram Eldar, Nov 26 2020 *)
-
for(n=0,12,for(k=0,n,print1((k^2*(n-k)^2+k*(k+2)*(n-k)+2)/2,", "));print(" "))
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