cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 19 results. Next

A121037 Multiples of 17 containing a 17 in their decimal representation.

Original entry on oeis.org

17, 170, 1173, 1700, 1717, 1734, 1751, 1768, 1785, 2176, 3179, 3417, 5117, 6171, 6817, 7174, 8177, 8517, 10217, 11713, 11730, 11747, 11764, 11781, 11798, 11917, 12172, 13175, 13617, 14178, 15317, 17000, 17017, 17034, 17051, 17068, 17085
Offset: 1

Views

Author

Reinhard Zumkeller, Jul 21 2006

Keywords

Crossrefs

Programs

  • Mathematica
    Select[17*Range[2000], StringContainsQ[IntegerString[#], "17"] &] (* Paolo Xausa, Feb 25 2024 *)
  • PARI
    is(n)=if(n%17, return(0)); while(n>16, if(n%100==17, return(1)); n\=10); 0 \\ Charles R Greathouse IV, Feb 12 2017

Formula

a(n) ~ 17n. - Charles R Greathouse IV, Feb 12 2017

Extensions

Corrected by T. D. Noe, Oct 25 2006

A033029 Numbers whose base-16 expansion has no run of digits with length < 2.

Original entry on oeis.org

17, 34, 51, 68, 85, 102, 119, 136, 153, 170, 187, 204, 221, 238, 255, 273, 546, 819, 1092, 1365, 1638, 1911, 2184, 2457, 2730, 3003, 3276, 3549, 3822, 4095, 4352, 4369, 4386, 4403, 4420, 4437, 4454, 4471, 4488, 4505, 4522, 4539
Offset: 1

Views

Author

Keywords

Comments

First term that is not a multiple of 17 (A008599) is 273, which is one more than 272 = 16 * 17. In hexadecimal, 272 is 110, which having a single 0 next to a 1, disqualifies it from this sequence. - Alonso del Arte, Oct 16 2016

Examples

			255 in hexadecimal is FF, a run of two Fs, hence 255 is in the sequence.
273 in hexadecimal is 111, a run of three 1s, hence 273 is in the sequence.
291 in hexadecimal is 123, three distinct single digits, so 291 is not in the sequence.
		

Programs

  • Mathematica
    Select[Range[10000], Min[Length/@Split[IntegerDigits[#, 16]]] > 1 &] (* Vincenzo Librandi, Feb 05 2014 *)

A141419 Triangle read by rows: T(n, k) = A000217(n) - A000217(n - k) with 1 <= k <= n.

Original entry on oeis.org

1, 2, 3, 3, 5, 6, 4, 7, 9, 10, 5, 9, 12, 14, 15, 6, 11, 15, 18, 20, 21, 7, 13, 18, 22, 25, 27, 28, 8, 15, 21, 26, 30, 33, 35, 36, 9, 17, 24, 30, 35, 39, 42, 44, 45, 10, 19, 27, 34, 40, 45, 49, 52, 54, 55
Offset: 1

Views

Author

Roger L. Bagula, Aug 05 2008

Keywords

Comments

As a rectangle, the accumulation array of A051340.
From Clark Kimberling, Feb 05 2011: (Start)
Here all the weights are divided by two where they aren't in Cahn.
As a rectangle, A141419 is in the accumulation chain
... < A051340 < A141419 < A185874 < A185875 < A185876 < ...
(See A144112 for the definition of accumulation array.)
row 1: A000027
col 1: A000217
diag (1,5,...): A000326 (pentagonal numbers)
diag (2,7,...): A005449 (second pentagonal numbers)
diag (3,9,...): A045943 (triangular matchstick numbers)
diag (4,11,...): A115067
diag (5,13,...): A140090
diag (6,15,...): A140091
diag (7,17,...): A059845
diag (8,19,...): A140672
(End)
Let N=2*n+1 and k=1,2,...,n. Let A_{N,n-1} = [0,...,0,1; 0,...,0,1,1; ...; 0,1,...,1; 1,...,1], an n X n unit-primitive matrix (see [Jeffery]). Let M_n=[A_{N,n-1}]^4. Then t(n,k)=[M_n](1,k), that is, the n-th row of the triangle is given by the first row of M_n. - _L. Edson Jeffery, Nov 20 2011
Conjecture. Let N=2*n+1 and k=1,...,n. Let A_{N,0}, A_{N,1}, ..., A_{N,n-1} be the n X n unit-primitive matrices (again see [Jeffery]) associated with N, and define the Chebyshev polynomials of the second kind by the recurrence U_0(x) = 1, U_1(x) = 2*x and U_r(x) = 2*x*U_(r-1)(x) - U_(r-2)(x) (r>1). Define the column vectors V_(k-1) = (U_(k-1)(cos(Pi/N)), U_(k-1)(cos(3*Pi/N)), ..., U_(k-1)(cos((2*n-1)*Pi/N)))^T, where T denotes matrix transpose. Let S_N = [V_0, V_1, ..., V_(n-1)] be the n X n matrix formed by taking V_(k-1) as column k-1. Let X_N = [S_N]^T*S_N, and let [X_N](i,j) denote the entry in row i and column j of X_N, i,j in {0,...,n-1}. Then t(n,k) = [X_N](k-1,k-1), and row n of the triangle is given by the main diagonal entries of X_N. Remarks: Hence t(n,k) is the sum of squares t(n,k) = sum[m=1,...,n (U_(k-1)(cos((2*m-1)*Pi/N)))^2]. Finally, this sequence is related to A057059, since X_N = [sum_{m=1,...,n} A057059(n,m)*A_{N,m-1}] is also an integral linear combination of unit-primitive matrices from the N-th set. - L. Edson Jeffery, Jan 20 2012
Row sums: n*(n+1)*(2*n+1)/6. - L. Edson Jeffery, Jan 25 2013
n-th row = partial sums of n-th row of A004736. - Reinhard Zumkeller, Aug 04 2014
T(n,k) is the number of distinct sums made by at most k elements in {1, 2, ... n}, for 1 <= k <= n, e.g., T(6,2) = the number of distinct sums made by at most 2 elements in {1,2,3,4,5,6}. The sums range from 1, to 5+6=11. So there are 11 distinct sums. - Derek Orr, Nov 26 2014
A number n occurs in this sequence A001227(n) times, the number of odd divisors of n, see A209260. - Hartmut F. W. Hoft, Apr 14 2016
Conjecture: 2*n + 1 is composite if and only if gcd(t(n,m),m) != 1, for some m. - L. Edson Jeffery, Jan 30 2018
From Peter Munn, Aug 21 2019 in respect of the sequence read as a triangle: (Start)
A number m can be found in column k if and only if A286013(m, k) is nonzero, in which case m occurs in column k on row A286013(m, k).
The first occurrence of m is in row A212652(m) column A109814(m), which is the rightmost column in which m occurs. This occurrence determines where m appears in A209260. The last occurrence of m is in row m column 1.
Viewed as a sequence of rows, consider the subsequences (of rows) that contain every positive integer. The lexicographically latest of these subsequences consists of the rows with row numbers in A270877; this is the only one that contains its own row numbers only once.
(End)

Examples

			As a triangle:
   1,
   2,  3,
   3,  5,  6,
   4,  7,  9, 10,
   5,  9, 12, 14, 15,
   6, 11, 15, 18, 20, 21,
   7, 13, 18, 22, 25, 27, 28,
   8, 15, 21, 26, 30, 33, 35, 36,
   9, 17, 24, 30, 35, 39, 42, 44, 45,
  10, 19, 27, 34, 40, 45, 49, 52, 54, 55;
As a rectangle:
   1   2   3   4   5   6   7   8   9  10
   3   5   7   9  11  13  15  17  19  21
   6   9  12  15  18  21  24  27  30  33
  10  14  18  22  26  30  34  38  42  46
  15  20  25  30  35  40  45  50  55  60
  21  27  33  39  45  51  57  63  69  75
  28  35  42  49  56  63  70  77  84  91
  36  44  52  60  68  76  84  92 100 108
  45  54  63  72  81  90  99 108 117 126
  55  65  75  85  95 105 115 125 135 145
Since the odd divisors of 15 are 1, 3, 5 and 15, number 15 appears four times in the triangle at t(3+(5-1)/2, 5) in column 5 since 5+1 <= 2*3, t(5+(3-1)/2, 3), t(1+(15-1)/2, 2*1) in column 2 since 15+1 > 2*1, and t(15+(1-1)/2, 1). - _Hartmut F. W. Hoft_, Apr 14 2016
		

References

  • R. N. Cahn, Semi-Simple Lie Algebras and Their Representations, Dover, NY, 2006, ISBN 0-486-44999-8, p. 139.

Crossrefs

Cf. A000330 (row sums), A004736, A057059, A070543.
A144112, A051340, A141419, A185874, A185875, A185876 are accumulation chain related.
A141418 is a variant.
Cf. A001227, A209260. - Hartmut F. W. Hoft, Apr 14 2016
A109814, A212652, A270877, A286013 relate to where each natural number appears in this sequence.
A000027, A000217, A000326, A005449, A045943, A059845, A115067, A140090, A140091, A140672 are rows, columns or diagonals - refer to comments.

Programs

  • Haskell
    a141419 n k =  k * (2 * n - k + 1) `div` 2
    a141419_row n = a141419_tabl !! (n-1)
    a141419_tabl = map (scanl1 (+)) a004736_tabl
    -- Reinhard Zumkeller, Aug 04 2014
  • Maple
    a:=(n,k)->k*n-binomial(k,2): seq(seq(a(n,k),k=1..n),n=1..12); # Muniru A Asiru, Oct 14 2018
  • Mathematica
    T[n_, m_] = m*(2*n - m + 1)/2; a = Table[Table[T[n, m], {m, 1, n}], {n, 1, 10}]; Flatten[a]

Formula

t(n,m) = m*(2*n - m + 1)/2.
t(n,m) = A000217(n) - A000217(n-m). - L. Edson Jeffery, Jan 16 2013
Let v = d*h with h odd be an integer factorization, then v = t(d+(h-1)/2, h) if h+1 <= 2*d, and v = t(d+(h-1)/2, 2*d) if h+1 > 2*d; see A209260. - Hartmut F. W. Hoft, Apr 14 2016
G.f.: y*(-x + y)/((-1 + x)^2*(-1 + y)^3). - Stefano Spezia, Oct 14 2018
T(n, 2) = A060747(n) for n > 1. T(n, 3) = A008585(n - 1) for n > 2. T(n, 4) = A016825(n - 2) for n > 3. T(n, 5) = A008587(n - 2) for n > 4. T(n, 6) = A016945(n - 3) for n > 5. T(n, 7) = A008589(n - 3) for n > 6. T(n, 8) = A017113(n - 4) for n > 7.r n > 5. T(n, 7) = A008589(n - 3) for n > 6. T(n, 8) = A017113(n - 4) for n > 7. T(n, 9) = A008591(n - 4) for n > 8. T(n, 10) = A017329(n - 5) for n > 9. T(n, 11) = A008593(n - 5) for n > 10. T(n, 12) = A017593(n - 6) for n > 11. T(n, 13) = A008595(n - 6) for n > 12. T(n, 14) = A147587(n - 7) for n > 13. T(n, 15) = A008597(n - 7) for n > 14. T(n, 16) = A051062(n - 8) for n > 15. T(n, 17) = A008599(n - 8) for n > 16. - Stefano Spezia, Oct 14 2018
T(2*n-k, k) = A070543(n, k). - Peter Munn, Aug 21 2019

Extensions

Simpler name by Stefano Spezia, Oct 14 2018

A008601 Multiples of 19.

Original entry on oeis.org

0, 19, 38, 57, 76, 95, 114, 133, 152, 171, 190, 209, 228, 247, 266, 285, 304, 323, 342, 361, 380, 399, 418, 437, 456, 475, 494, 513, 532, 551, 570, 589, 608, 627, 646, 665, 684, 703, 722, 741, 760, 779, 798, 817, 836, 855, 874, 893, 912, 931, 950, 969, 988
Offset: 0

Views

Author

Keywords

Crossrefs

Programs

Formula

(floor(a(n)/10) + 2*(a(n) mod 10)) == 0 modulo 19, see A076312. - Reinhard Zumkeller, Oct 06 2002
From Vincenzo Librandi, Dec 24 2010: (Start)
a(n) = 19*n.
a(n) = 2*a(n-1) - a(n-2).
G.f.: 19*x/(x-1)^2. (End)
From Elmo R. Oliveira, Apr 10 2025: (Start)
E.g.f.: 19*x*exp(x).
a(n) = (A008600(n) + A008602(n))/2. (End)

A244630 a(n) = 17*n^2.

Original entry on oeis.org

0, 17, 68, 153, 272, 425, 612, 833, 1088, 1377, 1700, 2057, 2448, 2873, 3332, 3825, 4352, 4913, 5508, 6137, 6800, 7497, 8228, 8993, 9792, 10625, 11492, 12393, 13328, 14297, 15300, 16337, 17408, 18513, 19652, 20825, 22032, 23273, 24548, 25857, 27200, 28577, 29988
Offset: 0

Views

Author

Vincenzo Librandi, Jul 03 2014

Keywords

Comments

First bisection of A195047. - Bruno Berselli, Jul 03 2014
Norms of purely imaginary numbers in Z[sqrt(-17)] (for example, 3*sqrt(-17) has norm 153). - Alonso del Arte, Jun 23 2018

Crossrefs

Cf. similar sequences of the type k*n^2: A000290 (k = 1), A001105 (k = 2), A033428 (k = 3), A016742 (k = 4), A033429 (k = 5), A033581 (k = 6), A033582 (k = 7), A139098 (k = 8), A016766 (k = 9), A033583 (k = 10), A033584 (k = 11), A135453 (k = 12), A152742 (k = 13), A144555 (k = 14), A064761 (k = 15), A016802 (k = 16), this sequence (k = 17), A195321 (k = 18), A244631 (k = 19), A195322 (k = 20), A064762 (k = 21), A195323 (k = 22), A244632 (k = 23), A195824 (k = 24), A016850 (k = 25), A244633 (k = 26), A244634 (k = 27), A064763 (k = 28), A244635 (k = 29), A244636 (k = 30).

Programs

Formula

G.f.: 17*x*(1 + x)/(1 - x)^3. [corrected by Bruno Berselli, Jul 03 2014]
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2.
a(n) = 17*A000290(n). - Omar E. Pol, Jul 03 2014
a(n) = a(-n). - Muniru A Asiru, Jun 29 2018
From Elmo R. Oliveira, Dec 02 2024: (Start)
E.g.f.: 17*x*(1 + x)*exp(x).
a(n) = n*A008599(n) = A195047(2*n). (End)

A094053 Triangle read by rows: T(n,k) = k*(n-k), 1 <= k <= n.

Original entry on oeis.org

0, 1, 0, 2, 2, 0, 3, 4, 3, 0, 4, 6, 6, 4, 0, 5, 8, 9, 8, 5, 0, 6, 10, 12, 12, 10, 6, 0, 7, 12, 15, 16, 15, 12, 7, 0, 8, 14, 18, 20, 20, 18, 14, 8, 0, 9, 16, 21, 24, 25, 24, 21, 16, 9, 0, 10, 18, 24, 28, 30, 30, 28, 24, 18, 10, 0, 11, 20, 27, 32, 35, 36, 35, 32, 27, 20, 11, 0, 12
Offset: 1

Views

Author

Reinhard Zumkeller, May 31 2004

Keywords

Comments

T(n,k) = A003991(n-1,k) for 1 <= k < n;
T(n,k) = T(n,n-1-k) for k < n;
T(n,1) = n-1; T(n,n) = 0; T(n,2) = A005843(n-2) for n > 1;
T(n,3) = A008585(n-3) for n>2; T(n,4) = A008586(n-4) for n > 3;
T(n,5) = A008587(n-5) for n>4; T(n,6) = A008588(n-6) for n > 5;
T(n,7) = A008589(n-7) for n>6; T(n,8) = A008590(n-8) for n > 7;
T(n,9) = A008591(n-9) for n>8; T(n,10) = A008592(n-10) for n > 9;
T(n,11) = A008593(n-11) for n>10; T(n,12) = A008594(n-12) for n > 11;
T(n,13) = A008595(n-13) for n>12; T(n,14) = A008596(n-14) for n > 13;
T(n,15) = A008597(n-15) for n>14; T(n,16) = A008598(n-16) for n > 15;
T(n,17) = A008599(n-17) for n>16; T(n,18) = A008600(n-18) for n > 17;
T(n,19) = A008601(n-19) for n>18; T(n,20) = A008602(n-20) for n > 19;
Row sums give A000292; triangle sums give A000332;
All numbers m > 0 occur A000005(m) times;
A002378(n) = T(A005408(n),n+1) = n*(n+1).
k-th columns are arithmetic progressions with step k, starting with 0. If a zero is prefixed to the sequence, then we get a new table where the columns are again arithmetic progressions with step k, but starting with k, k=0,1,2,...: 1st column = (0,0,0,...), 2nd column = (1,2,3,...), 3rd column = (2,4,6,8,...), etc. - M. F. Hasler, Feb 02 2013
Construct the infinite-dimensional matrix representation of angular momentum operators (J_1,J_2,J_3) in the Jordan-Schwinger form (cf. Harter, Klee, Schwinger). The triangle terms T(n,k) = T(2j,j+m) satisfy: (1/2)T(2j,j+m)^(1/2) = = = i = -i . Matrices for J_1 and J_2 are sparse. These equalities determine the only nonzero entries. - Bradley Klee, Jan 29 2016
T(n+1,k+1) is the number of degrees of freedom of a k-dimensional affine subspace within an n-dimensional vector space. This is most readily interpreted geometrically: e.g. in 3 dimensions a line (1-dimensional subspace) has T(4,2) = 4 degrees of freedom and a plane has T(4,3) = 3. T(n+1,1) = n indicates that points in n dimensions have n degrees of freedom. T(n,n) = 0 for any n as all n-dimensional spaces in an n-dimensional space are equivalent. - Daniel Leary, Apr 29 2020

Examples

			From _M. F. Hasler_, Feb 02 2013: (Start)
Triangle begins:
  0;
  1, 0;
  2, 2, 0;
  3, 4, 3, 0;
  4, 6, 6, 4, 0;
  5, 8, 9, 8, 5, 0;
  (...)
If an additional 0 was added at the beginning, this would become:
  0;
  0, 1;
  0, 2, 2;
  0, 3, 4; 3;
  0, 4, 6, 6, 4;
  0, 5, 8, 9, 8, 5;
  ... (End)
		

Crossrefs

J_3: A114327; J_1^2, J_2^2: A141387, A268759.
Cf. A000292 (row sums), A000332 (triangle sums).
T(n,k) for values of k:
A005843 (k=2), A008585 (k=3), A008586 (k=4), A008587 (k=5), A008588 (k=6), A008589 (k=7), A008590 (k=8), A008591 (k=9), A008592 (k=10), A008593 (k=11), A008594 (k=12), A008595 (k=13), A008596 (k=14), A008597 (k=15), A008598 (k=16), A008599 (k=17), A008600 (k=18), A008601 (k=19), A008602 (k=20).

Programs

  • Magma
    /* As triangle */ [[k*(n-k): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Jan 30 2016
    
  • Mathematica
    Flatten[Table[(j - m) (j + m + 1), {j, 0, 10, 1/2}, {m, -j, j}]] (* Bradley Klee, Jan 29 2016 *)
  • PARI
    {for(n=1, 13, for(k=1, n, print1(k*(n - k)," ");); print(););} \\ Indranil Ghosh, Mar 12 2017

A076311 a(n) = floor(n/10) - 5*(n mod 10).

Original entry on oeis.org

0, -5, -10, -15, -20, -25, -30, -35, -40, -45, 1, -4, -9, -14, -19, -24, -29, -34, -39, -44, 2, -3, -8, -13, -18, -23, -28, -33, -38, -43, 3, -2, -7, -12, -17, -22, -27, -32, -37, -42, 4, -1, -6, -11, -16, -21, -26, -31, -36, -41, 5, 0, -5, -10, -15, -20, -25, -30, -35
Offset: 0

Views

Author

Reinhard Zumkeller, Oct 06 2002

Keywords

Comments

(n==0 modulo 17) iff (a(n)==0 modulo 17); applied recursively, this property provides a divisibility test for numbers given in base 10 notation.

Examples

			12808 is not a multiple of 17, as 12808 -> 1280-5*8=1240 -> 124-5*0=124 -> 12-5*4=-8=17*(-1)+9, therefore the answer is NO.
Is 9248 divisible by 17? 9248 -> 924-5*8=884 -> 88-5*4=68=17*4, therefore the answer is YES.
		

References

  • Karl Menninger, Rechenkniffe, Vandenhoeck & Ruprecht in Goettingen (1961), 79A.

Crossrefs

Programs

  • Haskell
    a076311 n =  n' - 5 * m where (n', m) = divMod n 10
    -- Reinhard Zumkeller, Jun 01 2013
    
  • Magma
    [Floor(n/10)-5*(n mod 10): n in [0..50]]; // Vincenzo Librandi, Jun 23 2015
    
  • Mathematica
    Table[Floor[n/10]-5Mod[n,10],{n,0,60}] (* or *) LinearRecurrence[ {1,0,0,0,0,0,0,0,0,1,-1},{0,-5,-10,-15,-20,-25,-30,-35,-40,-45,1},60] (* Harvey P. Dale, Dec 21 2014 *)
  • PARI
    a(n)=n\10 - n%10*5 \\ Charles R Greathouse IV, Oct 07 2015

Formula

a(n)= +a(n-1) +a(n-10) -a(n-11). G.f. x *(-5-5*x-5*x^2-5*x^3-5*x^4-5*x^5-5*x^6-5*x^7-5*x^8+46*x^9) / ( (1+x) *(x^4+x^3+x^2+x+1) *(x^4-x^3+x^2-x+1) *(x-1)^2 ). - R. J. Mathar, Feb 20 2011

A195036 Vertex number of a square spiral in which the length of the first two edges are the legs of the primitive Pythagorean triple [15, 8, 17]. The edges of the spiral have length A195035.

Original entry on oeis.org

0, 15, 23, 53, 69, 114, 138, 198, 230, 305, 345, 435, 483, 588, 644, 764, 828, 963, 1035, 1185, 1265, 1430, 1518, 1698, 1794, 1989, 2093, 2303, 2415, 2640, 2760, 3000, 3128, 3383, 3519, 3789, 3933, 4218, 4370, 4670, 4830, 5145, 5313, 5643, 5819, 6164, 6348
Offset: 0

Views

Author

Omar E. Pol, Sep 12 2011

Keywords

Comments

Zero together with partial sums of A195035.
The only primes in the sequence are 23 and 53 since a(n) = (1/2)*((2*n+(-1)^n+3)/4)*((46*n-37*(-1)^n+37)/4). - Bruno Berselli, Sep 30 2011
The spiral contains infinitely many Pythagorean triples in which the hypotenuses on the main diagonal are the positives multiples of 17 (Cf. A008599). The vertices on the main diagonal are the numbers A195039 = (15+8)*A000217 = 23*A000217, where both 15 and 8 are the first two edges in the spiral. The distance "a" between nearest edges that are perpendicular to the initial edge of the spiral is 15, while the distance "b" between nearest edges that are parallel to the initial edge is 8, so the distance "c" between nearest vertices on the same axis is 17 because from the Pythagorean theorem we can write c = (a^2+b^2)^(1/2) = sqrt(15^2+8^2) = sqrt(225+64) = sqrt(289) = 17. - Omar E. Pol, Oct 12 2011

Crossrefs

Programs

Formula

From Bruno Berselli, Sep 30 2011: (Start)
G.f.: x*(15+8*x)/((1+x)^2*(1-x)^3).
a(n) = (2*n*(23*n+53) - (14*n+37)*(-1)^n + 37)/16.
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5). (End)

Extensions

More terms from Bruno Berselli, Sep 30 2011

A139618 a(n) = 153*n + 18.

Original entry on oeis.org

18, 171, 324, 477, 630, 783, 936, 1089, 1242, 1395, 1548, 1701, 1854, 2007, 2160, 2313, 2466, 2619, 2772, 2925, 3078, 3231, 3384, 3537, 3690, 3843, 3996, 4149, 4302, 4455, 4608, 4761, 4914, 5067, 5220, 5373, 5526, 5679, 5832, 5985
Offset: 0

Views

Author

Omar E. Pol, May 21 2008

Keywords

Comments

Numbers of the 18th column of positive numbers in the square array of nonnegative and polygonal numbers A139600. Also, numbers of the 18th column in the square array A057145.

Crossrefs

Programs

Formula

From Elmo R. Oliveira, Apr 04 2024: (Start)
G.f.: 9*(2+15*x)/(x-1)^2.
E.g.f.: 9*exp(x)*(2 + 17*x).
a(n) = 9*(A008599(n) + 2).
a(n) = 2*a(n-1) - a(n-2) for n >= 2. (End)

A317317 Multiples of 17 and odd numbers interleaved.

Original entry on oeis.org

0, 1, 17, 3, 34, 5, 51, 7, 68, 9, 85, 11, 102, 13, 119, 15, 136, 17, 153, 19, 170, 21, 187, 23, 204, 25, 221, 27, 238, 29, 255, 31, 272, 33, 289, 35, 306, 37, 323, 39, 340, 41, 357, 43, 374, 45, 391, 47, 408, 49, 425, 51, 442, 53, 459, 55, 476, 57, 493, 59, 510, 61, 527, 63, 544, 65, 561, 67, 578, 69
Offset: 0

Views

Author

Omar E. Pol, Jul 25 2018

Keywords

Comments

Partial sums give the generalized 21-gonal numbers (A303298).
a(n) is also the length of the n-th line segment of the rectangular spiral whose vertices are the generalized 21-gonal numbers.

Crossrefs

Cf. A008599 and A005408 interleaved.
Column 17 of A195151.
Sequences whose partial sums give the generalized k-gonal numbers: A026741 (k=5), A001477 (k=6), zero together with A080512 (k=7), A022998 (k=8), A195140 (k=9), zero together with A165998 (k=10), A195159 (k=11), A195161 (k=12), A195312 (k=13), A195817 (k=14).
Cf. A303298.

Programs

  • Mathematica
    With[{nn=40},Riffle[17*Range[0,nn],2*Range[0,nn]+1]] (* or *) LinearRecurrence[ {0,2,0,-1},{0,1,17,3},80] (* Harvey P. Dale, Jun 06 2020 *)
  • PARI
    concat(0, Vec(x*(1 + 17*x + x^2) / ((1 - x)^2*(1 + x)^2) + O(x^60))) \\ Colin Barker, Jul 29 2018

Formula

a(2n) = 17*n, a(2n+1) = 2*n + 1.
From Colin Barker, Jul 29 2018: (Start)
G.f.: x*(1 + 17*x + x^2) / ((1 - x)^2*(1 + x)^2).
a(n) = 2*a(n-2) - a(n-4) for n>3. (End)
Multiplicative with a(2^e) = 17*2^(e-1), and a(p^e) = p^e for an odd prime p. - Amiram Eldar, Oct 14 2023
Dirichlet g.f.: zeta(s-1) * (1 + 15/2^s). - Amiram Eldar, Oct 25 2023
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