A093918 -id:A093918 - cp's OEIS frontend

A093915 Triangle with r-th row containing r consecutive integers that sum to the smallest possible proper multiple of A006003(r).

2, 7, 8, 9, 10, 11, 24, 25, 26, 27, 24, 25, 26, 27, 28, 53, 54, 55, 56, 57, 58, 47, 48, 49, 50, 51, 52, 53, 94, 95, 96, 97, 98, 99, 100, 101, 78, 79, 80, 81, 82, 83, 84, 85, 86, 147, 148, 149, 150, 151, 152, 153, 154, 155, 156, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 212, 213, 214, 215, 216, 217

Offset: 1

Amarnath Murthy, Apr 25 2004

The r-th row constructed as explained in the example starts with x=A093916(r), ends with x+r-1=A093918(r), and has its sum A093917(r) equal to the smallest proper multiple of A006003(r). There is a simple formula for A093917(r), which allows us to calculate A093915(n) directly. - M. F. Hasler, Apr 04 2009

			Given the triangle
1 . . . . with row sum S1 = 1 = A006003(1)
2,3 . . . with row sum S2 = 2+3 = 5 = A006003(2)
4,5,6 . . with row sum S3 = 4+5+6 = 15 = A006003(3), etc.,
the sequence is constructed as follows:
The first row below must be a proper (i.e., > 1) multiple of S1; the smallest possibility is [ 2 ].
The next row below must contain 2 consecutive integers with sum equal to a proper multiple of S2=5. It cannot be 10 (not the sum of 2 consecutive integers), but 15 = 7+8 is a possibility.
The third row [x,x+1,x+2] must sum to a multiple of S3=15, and 2*S3=30 is possible for x=9.
The 4th row [x,x+1,x+2,x+3] must have its sum 4x+6 equal to a multiple of S4=7+8+9+10=34, and x=24 gives the sum 102=3*34, while 2*34=68 can't be achieved for any integer x.
This gives:
2 . . . . . . . with row sum 2 = 2*S1
7,8 . . . . . . with row sum 7+8 = 15 = 3*S2
9,10,11 . . . . with row sum 9+10+11 = 30 = 2*S3
24,25,26,27 . . with row sum 24+25+26+27 = 102 = 3*S4.

Cf. A093916, A093917, A093918.

for(r=1,9, x=A093916(r)-1; for(c=1,r, print1(x+c,","))) /* M. F. Hasler, Apr 04 2009 */

Edited and extended (values beyond a(15), example, PARI code) by M. F. Hasler, Apr 04 2009

A093916 a(2k-1) = (2k-1)^2 + 2 - k, a(2k) = 6k^2 + 2 - k: First column of the triangle A093915.

Original entry on oeis.org

2, 7, 9, 24, 24, 53, 47, 94, 78, 147, 117, 212, 164, 289, 219, 378, 282, 479, 353, 592, 432, 717, 519, 854, 614, 1003, 717, 1164, 828, 1337, 947, 1522, 1074, 1719, 1209, 1928, 1352, 2149, 1503, 2382, 1662, 2627, 1829, 2884, 2004, 3153, 2187, 3434, 2378, 3727, 2577, 4032, 2784, 4349, 2999, 4678, 3222, 5019, 3453, 5372

Offset: 1

Amarnath Murthy, Apr 25 2004

The sequence was initially defined as the first column of the triangle A093915, constructed by trial and error. It is however easy to prove that the sum of the r-th row of A093915, A093917(r), equals twice A006003(r) when r is odd, and three times A006003(r) when r is even. Given the expression of the row sum A093917(r) in terms of the first element a(r), one obtains the explicit formula for a(r). - M. F. Hasler, Apr 04 2009

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Cf. A093915, A093917, A093918.

[(n*(5*n-2) + (-1)^n*(n^2+1) + 7)/4: n in [1..70]]; // G. C. Greubel, Dec 30 2021

LinearRecurrence[{0,3,0,-3,0,1},{2,7,9,24,24,53},80] (* Harvey P. Dale, Nov 24 2017 *)

A093916(n)=((n^2+1)*(3-n%2)-n+1)/2
/* or the "experimental" version, trying out all allowed values */
A093916(n)={ local( s=(n^3+n)/2, d=(n^2-n)/2, k=ceil((2*s-d)/n)); while( (n*k+d)%s, k++ ); k } \\ M. F. Hasler, Apr 04 2009

[(5*n^2 -2*n +7 +(-1)^n*(n^2 +1))/4 for n in (1..70)] # G. C. Greubel, Dec 30 2021

a(n) = ((n^2+1)*b(n) - n + 1)/2 where b(n) = 3 - (n mod 2) = 2 if n odd, = 3 if n even. - M. F. Hasler, Apr 04 2009
From Colin Barker, May 01 2012: (Start)
a(n) = (n*(5*n-2) + (n^2+1)*(-1)^n + 7)/4.
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6).
G.f.: x*(2+7*x+3*x^2+3*x^3+3*x^4+2*x^5)/((1-x)^3*(1+x)^3). (End)
E.g.f.: (1/4)*( (7 +3*x +5*x^2)*exp(x) - 8 + (1 -x +x^2)*exp(-x) ). - G. C. Greubel, Dec 30 2021

Edited and extended by M. F. Hasler, Apr 04 2009

A093917 a(n) = n^3+n for odd n, (n^3+n)*3/2 for even n: Row sums of A093915.

Original entry on oeis.org

2, 15, 30, 102, 130, 333, 350, 780, 738, 1515, 1342, 2610, 2210, 4137, 3390, 6168, 4930, 8775, 6878, 12030, 9282, 16005, 12190, 20772, 15650, 26403, 19710, 32970, 24418, 40545, 29822, 49200, 35970, 59007, 42910, 70038, 50690, 82365, 59358

Offset: 1

Amarnath Murthy, Apr 25 2004

Initially defined as sum of the n-th row of the triangle A093915, constructed by trial and error. Namely, this row should contain n consecutive integers [x,x+1,...,x+n-1], listed in A093915, and have its sum a(n) = n*x+n(n-1)/2 equal to the least possible strict (>1) multiple of the sum of the indices of these elements in A093915, which equals A006003(n) = (n^3+n)/2. For odd n, a(n) = 2 A006003(n) is obtained for x = A093916(n). For even n, the sum a(n) cannot equal 2 A006003(n), but it does equal 3 A006003(n) for x = A093916(n). Hence this simple explicit definition of a(n). - M. F. Hasler, Apr 04 2009

Index entries for linear recurrences with constant coefficients, signature (0,4,0,-6,0,4,0,-1).

Cf. A093915, A093916, A093918.

a(n) = n*A093916(n)+n(n-1)/2. - M. F. Hasler, Apr 04 2009
a(2n-1) = 2*(2n-1)*(2n^2 -2n +1), a(2n) = 3*n*(4n^2 +1).
G.f.: x*(2+15*x+22*x^2+42*x^3+22*x^4+15*x^5+2*x^6) / ( (x-1)^4*(1+x)^4 ). - R. J. Mathar, Mar 21 2016

More terms from Jorge Coveiro, Jul 25 2006

Edited by M. F. Hasler, Apr 04 2009

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

Franck Maminirina Ramaharo, Apr 20 2018

Columns are linear recurrence sequences with signature (3,-3,1).
8*T(n,k) + A166147(k-1) are squares.
Columns k are binomial transforms of [1, k, 1, 0, 0, 0, ...].
Antidiagonals sums yield A116731.

			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.

Cf. A085475, A086270, A086271, A086272, A086273, A130154, A159798, A162609, A162610, A300401.

T := (n, k) -> binomial(n, 2) + k*n + 1;
for n from 0 to 20 do seq(T(n, k), k = 0 .. 20) od;

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));

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

G.f.: (3*x^2*y - 3*x*y + y - 2*x^2 + 2*x - 1)/((x - 1)^3*(y - 1)^2).
E.g.f.: (1/2)*(2*x*y + x^2 + 2)*exp(y + x).
T(n,k) = 3*T(n-1,k) - 3*T(n-2,k) + T(n-3,k), with T(0,k) = 1, T(1,k) = k + 1 and T(2,k) = 2*k + 2.
T(n,k) = T(n-1,k) + n + k - 1.
T(n,k) = T(n,k-1) + n, with T(n,0) = 1.
T(n,0) = A152947(n+1).
T(n,1) = A000124(n).
T(n,2) = A000217(n).
T(n,3) = A034856(n+1).
T(n,4) = A052905(n).
T(n,5) = A051936(n+4).
T(n,6) = A246172(n+1).
T(n,7) = A302537(n).
T(n,8) = A056121(n+1) + 1.
T(n,9) = A056126(n+1) + 1.
T(n,10) = A051942(n+10) + 1, n > 0.
T(n,11) = A101859(n) + 1.
T(n,12) = A132754(n+1) + 1.
T(n,13) = A132755(n+1) + 1.
T(n,14) = A132756(n+1) + 1.
T(n,15) = A132757(n+1) + 1.
T(n,16) = A132758(n+1) + 1.
T(n,17) = A212427(n+1) + 1.
T(n,18) = A212428(n+1) + 1.
T(n,n) = A143689(n) = A300192(n,2).
T(n,n+1) = A104249(n).
T(n,n+2) = T(n+1,n) = A005448(n+1).
T(n,n+3) = A000326(n+1).
T(n,n+4) = A095794(n+1).
T(n,n+5) = A133694(n+1).
T(n+2,n) = A005449(n+1).
T(n+3,n) = A115067(n+2).
T(n+4,n) = A133694(n+2).
T(2*n,n) = A054556(n+1).
T(2*n,n+1) = A054567(n+1).
T(2*n,n+2) = A033951(n).
T(2*n,n+3) = A001107(n+1).
T(2*n,n+4) = A186353(4*n+1) (conjectured).
T(2*n,n+5) = A184103(8*n+1) (conjectured).
T(2*n,n+6) = A250657(n-1) = A250656(3,n-1), n > 1.
T(n,2*n) = A140066(n+1).
T(n+1,2*n) = A005891(n).
T(n+2,2*n) = A249013(5*n+4) (conjectured).
T(n+3,2*n) = A186384(5*n+3) = A186386(5*n+3) (conjectured).
T(2*n,2*n) = A143689(2*n).
T(2*n+1,2*n+1) = A143689(2*n+1) (= A030503(3*n+3) (conjectured)).
T(2*n,2*n+1) = A104249(2*n) = A093918(2*n+2) = A131355(4*n+1) (= A030503(3*n+5) (conjectured)).
T(2*n+1,2*n) = A085473(n).
a(n+1,5*n+1)=A051865(n+1) + 1.
a(n,2*n+1) = A116668(n).
a(2*n+1,n) = A054569(n+1).
T(3*n,n) = A025742(3*n-1), n > 1 (conjectured).
T(n,3*n) = A140063(n+1).
T(n+1,3*n) = A069099(n+1).
T(n,4*n) = A276819(n).
T(4*n,n) = A154106(n-1), n > 0.
T(2^n,2) = A028401(n+2).
T(1,n)*T(n,1) = A006000(n).
T(n*(n+1),n) = A211905(n+1), n > 0 (conjectured).
T(n*(n+1)+1,n) = A294259(n+1).
T(n,n^2+1) = A081423(n).
T(n,A000217(n)) = A158842(n), n > 0.
T(n,A152947(n+1)) = A060354(n+1).
floor(T(n,n/2)) = A267682(n) (conjectured).
floor(T(n,n/3)) = A025742(n-1), n > 0 (conjectured).
floor(T(n,n/4)) = A263807(n-1), n > 0 (conjectured).
ceiling(T(n,2^n)/n) = A134522(n), n > 0 (conjectured).
ceiling(T(n,n/2+n)/n) = A051755(n+1) (conjectured).
floor(T(n,n)/n) = A133223(n), n > 0 (conjectured).
ceiling(T(n,n)/n) = A007494(n), n > 0.
ceiling(T(n,n^2)/n) = A171769(n), n > 0.
ceiling(T(2*n,n^2)/n) = A046092(n), n > 0.
ceiling(T(2*n,2^n)/n) = A131520(n+2), n > 0.

cp's OEIS Frontend

A093915 Triangle with r-th row containing r consecutive integers that sum to the smallest possible proper multiple of A006003(r).

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A093916 a(2*k-1) = (2*k-1)^2 + 2 - k, a(2*k) = 6*k^2 + 2 - k: First column of the triangle A093915.

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A093917 a(n) = n^3+n for odd n, (n^3+n)*3/2 for even n: Row sums of A093915.

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A303273 Array T(n,k) = binomial(n, 2) + k*n + 1 read by antidiagonals.

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A093916 a(2k-1) = (2k-1)^2 + 2 - k, a(2k) = 6k^2 + 2 - k: First column of the triangle A093915.