A017377 -id:A017377 - cp's OEIS frontend

A130423 Main diagonal of array A[k,n] = n-th sum of 3 consecutive k-gonal numbers, k>2.

4, 14, 39, 88, 170, 294, 469, 704, 1008, 1390, 1859, 2424, 3094, 3878, 4785, 5824, 7004, 8334, 9823, 11480, 13314, 15334, 17549, 19968, 22600, 25454, 28539, 31864, 35438, 39270, 43369, 47744, 52404, 57358, 62615, 68184, 74074, 80294, 86853

Offset: 1

Jonathan Vos Post, May 26 2007

The first row of the array is the sum of 3 consecutive triangular numbers = A000217(n) + A000217(n+1) + A000217(n+2) = Centered triangular numbers: 3*n*(n-1)/2 + 1, for n>1. The second row of the array is the sum of 3 consecutive squares = Number of points on surface of square pyramid: 3*n^2 + 2 (n>1). The first column of the array is k+1 = 4, 5, 6, 7, 8, 9, ... The second column of the array is A016825 = 4*n + 2 (for n>2). The third column of the array is A017377 = 10*n + 9 (for n>0).

			The array begins:
k / A[k,n]
3.|.4.10.19.31..46..64..85.109.136.166....=A005448(n+1).
4.|.5.14.29..50..77.110.149.194.245.302...=A005918(n).
5.|.6.18.39..69.108.156.213.279.354.438...=A129863(n).
6.|.7.22.49..88.139.202.277.364.463.574...
7.|.8.26.59.107.170.248.341.449.572.710...
8.|.9.30.69.126.201.294.405.534.681.846...

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Polygonal Number
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000217, A000290, A000326, A000384, A000566, A000567, A005448, A005918, A016825, A017377, A129803, A129863.

I:=[4, 14, 39, 88]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Jun 28 2012

P := proc(k,n) n*((k-2)*n-k+4)/2 ; end: A := proc(k,n) add( P(k,i),i=n..n+2) ; end: A130423 := proc(n) A(n+3,n) ; end: seq(A130423(n),n=0..40) ; # R. J. Mathar, Jun 14 2007

CoefficientList[Series[(4-2*x+7*x^2)/(1-x)^4,{x,0,40}],x] (* Vincenzo Librandi, Jun 28 2012 *)
Table[n (3n^2-3n+8)/2,{n,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{4,14,39,88},40] (* Harvey P. Dale, Aug 15 2012 *)

a(n) = A[n+2,n] = P(k+2,n) + P(k+2,n+1) + P(k+2,n+2) where P(k,n) = k*((n-2)*k - (n-4))/2.
a(n) = n*(3*n^2-3*n+8)/2. G.f.: x*(4-2*x+7*x^2)/(1-x)^4. [Colin Barker, Apr 30 2012]
a(1)=4, a(2)=14, a(3)=39, a(4)=88, a(n)=4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - Harvey P. Dale, Aug 15 2012

More terms from R. J. Mathar, Jun 14 2007

A190876 Numbers 1 through 8 together with numbers congruent to 9 mod 10.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 19, 29, 39, 49, 59, 69, 79, 89, 99, 109, 119, 129, 139, 149, 159, 169, 179, 189, 199, 209, 219, 229, 239, 249, 259, 269, 279, 289, 299, 309, 319, 329, 339, 349, 359, 369, 379, 389, 399, 409, 419, 429, 439, 449, 459, 469, 479, 489, 499, 509, 519, 529, 539, 549, 559, 569, 579, 589, 599, 609

Offset: 1

N. J. A. Sloane, May 23 2011

In lunar arithmetic, numbers n with the property that the sum of the divisors of n that are <= n is equal to n.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
D. Applegate, M. LeBrun and N. J. A. Sloane, Dismal Arithmetic [Note: we have now changed the name from "dismal arithmetic" to "lunar arithmetic" - the old name was too depressing]
Index entries for sequences related to dismal (or lunar) arithmetic
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A017377.

[n lt 9 select n else 10*n-81: n in [1..70]]; // Bruno Berselli, May 24 2011

CoefficientList[Series[x*(1+9*x^9)/(1-x)^2,{x,0,69}],x] (* Stefano Spezia, May 26 2025 *)

makelist(if n<9 then n else 10*n-81,n,1,70); /* Bruno Berselli, May 24 2011 */

G.f.: x*(1+9*x^9)/(1-x)^2. - Bruno Berselli, May 23 2011

a(n) = 2*a(n-1) -a(n-2). - Vincenzo Librandi, Jul 12 2012

A290972 Primes p such that the sum of the squares of digits of p equals the sum of digits of p^2.

Original entry on oeis.org

2, 3, 3331, 3433, 11243, 13241, 21523, 22153, 22531, 31541, 32141, 32411, 33203, 34033, 34141, 34211, 35141, 41341, 41413, 42131, 43411, 44131, 51341, 51413, 52321, 54311, 102253, 102523, 104231, 104513, 110543, 111263, 111623, 112163

Offset: 1

K. D. Bajpai, Aug 16 2017

214007 is the smallest term that is in A017353 and 31111009 is the smallest term that is in A017377. - Altug Alkan, Aug 16 2017

			a(3) = 3331 is prime: [3^2 + 3^2 + 3^2 + 1^2 = 9 + 9 + 9 + 1] = 28; [3331^2 = 11095561, 1 + 1 + 0 + 9 + 5 + 5 + 1] = 28.
a(5) = 11243 is prime: [1^2 + 1^2 + 2^2 + 4^2 + 3^2 = 1 + 1 + 4 + 16 + 9] = 31: [11243^2 = 126405049;1 + 2 + 6 + 4 + 0 + 5 + 0 + 4 + 9] = 31.

Robert Israel, Table of n, a(n) for n = 1..4000

Intersection of A000040 and A165550.

Cf. A123157.

filter:= t -> convert(map(`^`,convert(t,base,10),2),`+`) = convert(convert(t^2,base,10),`+`) and isprime(t):
select(filter, [2,seq(i,i=3..200000,2)]); # Robert Israel, Aug 16 2017

Select[Prime[Range[20000]], Plus @@ IntegerDigits[#^2] == Total[IntegerDigits[#]^2] &]

forprime(p=1, 30000, v=digits(p); if(sum(i=1, length(v), v[i]^2) == sumdigits(p^2), print1(p", ")));

A348491 Positive numbers whose square starts and ends with exactly one 9.

Original entry on oeis.org

3, 97, 303, 307, 313, 953, 957, 963, 967, 973, 977, 983, 987, 993, 3003, 3007, 3013, 3017, 3023, 3027, 3033, 3037, 3043, 3047, 3053, 3057, 3063, 3067, 3073, 3077, 3083, 3087, 3093, 3097, 3103, 3107, 3113, 3117, 3123, 3127, 3133, 3137, 3143, 9487, 9493, 9497, 9503, 9507, 9513, 9517

Offset: 1

Bernard Schott, Nov 02 2021

When a square ends with 9, it ends with only one 9.
From Marius A. Burtea, Nov 02 2021 : (Start)
The sequence is infinite because the numbers 303, 3003, 30003, ..., 3*(10^k + 1), k >= 2, are terms with squares 91809, 9018009, 900180009, 90001800009, ... 9*(10^(2*k) + 2*10^k + 1), k >= 2.
Numbers 97, 967, 9667, 96667, 966667, ..., (29*10^n + 1) / 3, k >= 1, are terms and have no digits 0, because their squares are 9409, 935089, 93450889, 9344508889, 934445088889, ...
Also 963, 9663, 96663, 966663, 9666663, 96666663, ... (29*10^k - 11) / 3, k >= 2, are terms and have no digits 0, because their squares are 927369, 93373569, 9343735569, 934437355569, 93444373555569, 9344443735555569, ... (End)

			97^2 = 9409, hence 97 is a term.
997^2 = 994009, hence  997 is not a term.

Subsequence of A305719, A063226, and A045863.
Cf. A017377, A045863, A273374 (squares ending with 9).
Similar to: A348487 (k=1), A348488 (k=4), A348489 (k=5), A348490 (k=6), this sequence (k=9).

[3] cat [n:n in [4..9600]|Intseq(n*n)[1] eq 9 and Intseq(n*n)[#Intseq(n*n)] eq 9]; // Marius A. Burtea, Nov 02 2021

Join[{3}, Select[Range[10, 10^4], (d = IntegerDigits[#^2])[[1]] == d[[-1]] == 9 && d[[2]] != 9 &]] (* Amiram Eldar, Nov 02 2021 *)

isok(k) = my(d=digits(sqr(k))); (d[1]==9) && (d[#d]==9) && if (#d>2, (d[2]!=9) && (d[#d-1]!=9), 1); \\ Michel Marcus, Nov 03 2021

list(lim)=my(v=List([3])); for(d=2, 2*#digits(lim\=1), my(s=sqrtint(9*10^(d-1)-1)+1); s+=[3,2,1,0,3,2,1,0,5,4][s%10+1]; forstep(n=s, min(sqrtint(10^d-10^(d-2)-1), lim), if(s%10==3, [4,6], [6,4]), listput(v, n))); Vec(v) \\ Charles R Greathouse IV, Nov 03 2021

from itertools import count, takewhile
def ok(n):
  s = str(n*n); return len(s.rstrip("9")) == len(s.lstrip("9")) == len(s)-1
def aupto(N):
  r = takewhile(lambda x: x<=N, (10*i+d for i in count(0) for d in [3, 7]))
  return [k for k in r if ok(k)]
print(aupto(9517)) # Michael S. Branicky, Nov 02 2021

A354337 a(n) is the integer w such that (L(2n)^2, -L(2n + 1)^2, w) is a primitive solution to the Diophantine equation 2x^3 + 2y^3 + z^3 = 125, where L(n) is the n-th Lucas number (A000032).

Original entry on oeis.org

19, 149, 1039, 7139, 48949, 335519, 2299699, 15762389, 108037039, 740496899, 5075441269, 34787591999, 238437702739, 1634276327189, 11201496587599, 76776199786019, 526231901914549, 3606847113615839, 24721697893396339, 169445038140158549, 1161393569087713519

Offset: 1

XU Pingya, Jun 20 2022

Subsequence of A017377.

			2*(L(4)^2)^3 + 2*(-L(5)^2)^3 + (149)^3 = 2*(49)^3 + 2*(-121)^3 + (149)^3 = 125, a(2) = 149.

Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000032, A002878, A005248, A017377, A081017, A089508, A092521, A288913.

Cf. A337929, A354336.

LinearRecurrence[{7,-1},{19,149},21]-1 + LucasL[2*Range[21]-3]^2

a(n) = (125 - 2*A005248(n)^6 + 2*A002878(n)^6)^(1/3).
a(n) = Lucas(4*n+2) + Lucas(4n-1) - 3 = 2*A056914(n)-3 = 15*A092521(n) + A288913(n-1).
a(n) = 2*A081017(n) - 1.
a(n) = 10*A089508(n) + 9.
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
G.f.: x*(19 - 3*x - x^2)/((1 - x)*(1 - 7*x + x^2)). - Stefano Spezia, Jun 22 2022

A373460 Numbers k such that k and k+1 both have an equal number of even and odd digits.

Original entry on oeis.org

29, 49, 69, 89, 1009, 1029, 1049, 1069, 1089, 1209, 1229, 1249, 1269, 1289, 1409, 1429, 1449, 1469, 1489, 1609, 1629, 1649, 1669, 1689, 1809, 1829, 1849, 1869, 1889, 2109, 2129, 2149, 2169, 2189, 2309, 2329, 2349, 2369, 2389, 2509, 2529, 2549, 2569, 2589, 2709

Offset: 1

Amiram Eldar, Jun 07 2024

The terms are of the form 100*m + j, where m is either 0 or a term of A227870 and j is in {29, 49, 69, 89} if m = 0 or in {9, 29, 49, 69, 89} if m > 0.

			29 is a term since it has one even digit (2) and one odd digit (9), and 29+1 = 30 also has one even digit (0) and one odd digit (3).

Amiram Eldar, Table of n, a(n) for n = 1..10000

Subsequence of A017377 and A227870.

Cf. A337238 (binary analog), A373505.

q[n_] := Module[{d = Differences[Tally[Mod[IntegerDigits[n], 2]]]}, d != {} && d[[1, 2]] == 0]; Select[Range[3000], q[#] && q[# + 1] &]

iseq(n) = {my(o = 0, e = 0); while(n > 0, if((n%10) % 2 == 0, e++, o++); n \= 10); e == o;}
lista(kmax) = {my(q1 = 0, q2); for(k = 1, kmax, q2 = iseq(k); if(q1 && q2, print1(k-1, ", ")); q1 = q2);}

a(n) = 100 * A227870(floor(n/5)) + 20 * (n mod 5) + 9, for n > 4.

A385623 Array read by ascending antidiagonals: A(n,k) is the number obtained by concatenation of n with k in that order, with k >= 0.

Original entry on oeis.org

0, 10, 1, 20, 11, 2, 30, 21, 12, 3, 40, 31, 22, 13, 4, 50, 41, 32, 23, 14, 5, 60, 51, 42, 33, 24, 15, 6, 70, 61, 52, 43, 34, 25, 16, 7, 80, 71, 62, 53, 44, 35, 26, 17, 8, 90, 81, 72, 63, 54, 45, 36, 27, 18, 9, 100, 91, 82, 73, 64, 55, 46, 37, 28, 19, 10, 110, 101, 92, 83, 74, 65, 56, 47, 38, 29, 110, 11

Offset: 0

Stefano Spezia, Jul 05 2025

			Array begins as:
   0,  1,  2,  3,  4,  5,  6,  7, ...
  10, 11, 12, 13, 14, 15, 16, 17, ...
  20, 21, 22, 23, 24, 25, 26, 27, ...
  30, 31, 32, 33, 34, 35, 36, 37, ...
  40, 41, 42, 43, 44, 45, 46, 47, ...
  50, 51, 52, 53, 54, 55, 56, 57, ...
  60, 61, 62, 63, 64, 65, 66, 67, ...
  ...

Stefano Spezia, Table of n, a(n) for n = 0..11475 (first 151 antidiagonals of the array, flattened)

Columns for k=0..9 give: A008592, A017281, A017293, A017305, A017317, A017329, A017341, A017353, A017365, A017377.

Cf. A001477 (1st row), A020338 (main diagonal), A055642, A385624 (antidiagonal sums).

A[n_,k_]:=FromDigits[Join[IntegerDigits[n],IntegerDigits[k]]]; Table[A[n,k],{n,0,6},{k,0,7}] (* or *)
A[n_,k_]:=If[k==0,10n,n*10^(Floor[Log10[k]]+1)+k]; Table[A[n-k,k],{n,0,11},{k,0,n}]//Flatten

T(n, k) = fromdigits(concat(digits(n), digits(k))); \\ Michel Marcus, Jul 06 2025

A(n,0) = 10*n and A(n,k) = n*10^(floor(log_10(k)) + 1) + k for k > 0.

A001535 a(n) = (10n+1)*(10n+9).

Original entry on oeis.org

9, 209, 609, 1209, 2009, 3009, 4209, 5609, 7209, 9009, 11009, 13209, 15609, 18209, 21009, 24009, 27209, 30609, 34209, 38009, 42009, 46209, 50609, 55209, 60009, 65009, 70209, 75609, 81209, 87009, 93009, 99209, 105609, 112209, 119009, 126009, 133209, 140609

Offset: 0

N. J. A. Sloane

T. D. Noe, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001622, A017281, A017377.

seq((10*n+1)*(10*n+9),n = 0 .. 100); # Robert Israel, Dec 17 2014

Times@@@Table[10n+{1,9},{n,0,40}] (* or *) LinearRecurrence[{3,-3,1},{9,209,609},40] (* Harvey P. Dale, Oct 15 2014 *)
CoefficientList[Series[(9 + 182 x + 9 x^2) / (1 - x)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Dec 17 2014 *)

a(n)=(10*n+1)*(10*n+9) \\ Charles R Greathouse IV, Jun 16 2017

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=9, a(1)=209, a(2)=609. - Harvey P. Dale, Oct 15 2014
G.f.: (9 + 182*x + 9*x^2)/(1 - x)^3. - Vincenzo Librandi, Dec 17 2014
E.g.f.: (100*x^2 + 200*x + 9)*exp(x). - Robert Israel, Dec 17 2014
From Amiram Eldar, Feb 20 2023: (Start)
a(n) = A017281(n)*A017377(n).
Sum_{n>=0} 1/a(n) = sqrt(5+2*sqrt(5))*Pi/80.
Sum_{n>=0} (-1)^n/a(n) = (sqrt(10+2*sqrt(5)) * log(cot(Pi/20)) + sqrt(10-2*sqrt(5)) * log(cot(3*Pi/20)))/80.
Product_{n>=0} (1 - 1/a(n)) = 2*phi*cos(sqrt(17)*Pi/10), where phi is the golden ratio (A001622).
Product_{n>=0} (1 + 1/a(n)) = 2*phi*cos(sqrt(15)*Pi/10). (End)

A017384 a(n) = (10*n + 9)^8.

Original entry on oeis.org

43046721, 16983563041, 500246412961, 5352009260481, 33232930569601, 146830437604321, 513798374428641, 1517108809906561, 3936588805702081, 9227446944279201, 19925626416901921, 40213853471634241, 76686282021340161

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (9, -36, 84, -126, 126, -84, 36, -9, 1).

Cf. A001016, A017377.

[(10*n+9)^8: n in [0..20]]; // Vincenzo Librandi, Sep 01 2011

(10 Range[0, 19] + 9)^8 (* Alonso del Arte, Dec 02 2017 *)

a(n) = (10*n + 9)^8 \\ Iain Fox, Dec 02 2017

From Iain Fox, Dec 02 2017: (Start)
a(n) = 9*a(n-1) - 36*a(n-2) + 84*a(n-3) - 126*a(n-4) + 126*a(n-5) - 84*a(n-6) + 36*a(n-7) - 9*a(n-8) + a(n-9), n > 8.
G.f.: (43046721 + 16596142552*x + 348944027548*x^2 + 1457583888744*x^3 + 1652522683270*x^4 + 520202222824*x^5 + 35893629468*x^6 + 214358872*x^7 + x^8)/(1-x)^9.
E.g.f.: exp(x)*(43046721 + 16940516320*x + 233161166800*x^2 + 650362944000*x^3 + 614936700000*x^4 + 243902400000*x^5 + 43988000000*x^6 + 3520000000*x^7 + 100000000*x^8). (End)
a(n) = A001016(A017377(n)). - Felix Fröhlich, Dec 03 2017

A130424 Main diagonal of array A[k,n] = n-th sum of k consecutive k-gonal numbers, k>2.

Original entry on oeis.org

4, 30, 125, 365, 854, 1724, 3135, 5275, 8360, 12634, 18369, 25865, 35450, 47480, 62339, 80439, 102220, 128150, 158725, 194469, 235934, 283700, 338375, 400595, 471024, 550354, 639305, 738625, 849090, 971504, 1106699, 1255535, 1418900

Offset: 1

Jonathan Vos Post, May 26 2007

The first row of the array is the sum of 3 consecutive triangular numbers = A000217(n) + A000217(n+1) + A000217(n+2) = Centered triangular numbers: 3*n*(n-1)/2 + 1, for n>1. The second row of the array is the sum of 4 consecutive squares = n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 = A027575(n). The third row of the array is the sum of 5 consecutive pentagonal numbers.

			The array begins:
k / A[k,n]
3.|...4..10..19...31...46...64...85..109.136.166...=A005448(n+1).
4.|..14..30..54...86..126..174..230..294.366.446...=A027575(n).
5.|..40..75.125..190..270..365..475..600.740...
6.|..95.161.251..365..503..665..851.1061.1295...
7.|.196.308.455..637..854.1106.1393.1715.2072...
8.|.364.540.764.1036.1356.1724.2140.2604.3116...

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Polygonal Number.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000217, A000290, A000326, A000384, A000566, A000567, A005448, A005918, A016825, A017377, A129803, A129863, A130423.

P := proc(k,n) n*((k-2)*n-k+4)/2 ; end: A := proc(k,n) add( P(k,i),i=n..n+k-1) ; end: A130424 := proc(n) A(n+3,n) ; end: seq(A130424(n),n=0..40) ; # R. J. Mathar, Oct 28 2007

LinearRecurrence[{5,-10,10,-5,1},{4,30,125,365,854},50] (* Harvey P. Dale, Jun 23 2020 *)

a(n) = A[n+2,n] = P(k+2,n) + P(k+2,n+1) + P(k+2,n+2) + ... P(k+2,n+k-1) where P(k,n) = k*((n-2)*k - (n-4))/2.
a(n) = (n+2)*(7*n^3-8*n^2+12*n-3)/6. [R. J. Mathar, Oct 30 2008]
G.f.: x*(4+10*x+15*x^2-x^4)/(1-x)^5. [Colin Barker, Sep 08 2012]

More terms from R. J. Mathar, Oct 28 2007

cp's OEIS Frontend

A130423 Main diagonal of array A[k,n] = n-th sum of 3 consecutive k-gonal numbers, k>2.

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A190876 Numbers 1 through 8 together with numbers congruent to 9 mod 10.

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A290972 Primes p such that the sum of the squares of digits of p equals the sum of digits of p^2.

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A348491 Positive numbers whose square starts and ends with exactly one 9.

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A354337 a(n) is the integer w such that (L(2*n)^2, -L(2*n + 1)^2, w) is a primitive solution to the Diophantine equation 2*x^3 + 2*y^3 + z^3 = 125, where L(n) is the n-th Lucas number (A000032).

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A373460 Numbers k such that k and k+1 both have an equal number of even and odd digits.

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A385623 Array read by ascending antidiagonals: A(n,k) is the number obtained by concatenation of n with k in that order, with k >= 0.

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A354337 a(n) is the integer w such that (L(2n)^2, -L(2n + 1)^2, w) is a primitive solution to the Diophantine equation 2x^3 + 2y^3 + z^3 = 125, where L(n) is the n-th Lucas number (A000032).