A016897 -id:A016897 - cp's OEIS frontend

A120433 Numbers whose Roman numeral representation uses the subtractive notation.

4, 9, 14, 19, 24, 29, 34, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 54, 59, 64, 69, 74, 79, 84, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 104, 109, 114, 119, 124, 129, 134, 139, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149, 154, 159, 164, 169, 174

Offset: 1

Christian Amet, Jul 13 2006

Each number in this sequence has a 4 or a 9 in its decimal representation, corresponding to one of IV, IX, XL, XC, CD, CM. - Alonso del Arte, Jan 05 2018

			In Roman numerals, 14 is XIV, that is, X + (V - I) = 10 + (5 - 1) = 14, so 14 is in the sequence.
In Roman numerals, 15 is XV, meaning X + V = 10 + 5 = 15, which does not use subtractive notation, so 15 is not in the sequence.

Nathaniel Johnston, Table of n, a(n) for n = 1..1952 (complete up to 3999)

Cf. A061493.

Cf. A016897, 5n + 4 (first diverges after 39, as that sequence does not include 40, 41, 42, 43).

with(StringTools): for n from 1 to 300 do r:=convert(n,roman): if(Search("IV",r)>0 or Search("IX",r)>0 or Search("XL",r)>0 or Search("XC",r)>0 or Search("CD",r)>0 or Search("CM",r)>0)then printf("%d, ", n): fi: od: # Nathaniel Johnston, May 18 2011

Select[Range[3999], StringContainsQ[RomanNumeral[#], {"IV", "IX", "XL", "XC", "CD", "CM"}] &] (* Michael De Vlieger, Aug 20 2024 *)

def ok(n): return {"4", "9"} & set(str(n))
afull = [k for k in range(4000) if ok(k)] # Michael S. Branicky, Aug 20 2024

A144204 Array A(k,n) = (n+k-2)*(n-1) - 1 (k >= 1, n >= 1) read by antidiagonals.

Original entry on oeis.org

-1, -1, 0, -1, 1, 3, -1, 2, 5, 8, -1, 3, 7, 11, 15, -1, 4, 9, 14, 19, 24, -1, 5, 11, 17, 23, 29, 35, -1, 6, 13, 20, 27, 34, 41, 48, -1, 7, 15, 23, 31, 39, 47, 55, 63, -1, 8, 17, 26, 35, 44, 53, 62, 71, 80, -1, 9, 19, 29, 39, 49, 59, 69, 79, 89, 99, -1, 10, 21, 32, 43, 54, 65, 76, 87

Offset: 1

Jonathan Vos Post, Sep 13 2008

Arises in complete intersection threefolds,
Also can be produced as a triangle read by rows: a(n, k) = nk - (n + k). - Alonso del Arte, Jul 09 2009
Kosta: Let X be a complete intersection of two hypersurfaces F_n and F_k in the projective space P^5 of degree n and k respectively. with n=>k, such that the singularities of X are nodal and F_k is smooth. We prove that if the threefold X has at most (n+k-2)*(n-1) - 1 singular points, then it is factorial.

			From _R. J. Mathar_, Jul 10 2009: (Start)
The rows A(n,1), A(n,2), A(n,3), etc., are :
.-1...0...3...8..15..24..35..48..63..80..99.120.143.168 A067998
.-1...1...5..11..19..29..41..55..71..89.109.131.155.181 A028387
.-1...2...7..14..23..34..47..62..79..98.119.142.167.194 A008865
.-1...3...9..17..27..39..53..69..87.107.129.153.179.207 A014209
.-1...4..11..20..31..44..59..76..95.116.139.164.191.220 A028875
.-1...5..13..23..35..49..65..83.103.125.149.175.203.233 A108195
.-1...6..15..26..39..54..71..90.111.134.159.186.215.246
.-1...7..17..29..43..59..77..97.119.143.169.197.227.259
.-1...8..19..32..47..64..83.104.127.152.179.208.239.272
.-1...9..21..35..51..69..89.111.135.161.189.219.251.285
.-1..10..23..38..55..74..95.118.143.170.199.230.263.298
.-1..11..25..41..59..79.101.125.151.179.209.241.275.311
.-1..12..27..44..63..84.107.132.159.188.219.252.287.324
.-1..13..29..47..67..89.113.139.167.197.229.263.299.337 Cf. A126719.
(End)
As a triangle:
. 0
. 1, 3
. 2, 5, 8
. 3, 7, 11, 15
. 4, 9, 14, 19, 24
. 5, 11, 17, 23, 29, 35
. 6, 13, 20, 27, 34, 41, 48
. 7, 15, 23, 31, 39, 47, 55, 63
. 8, 17, 26, 35, 44, 53, 62, 71, 80

Dimitra Kosta, Factoriality of complete intersection threefolds arXiv:0808.4071 [math.AG]

Row 1 = A067998(n) for n>0. Row 2 = A028387(n) for n>0.Column 1 = -A000012(n). Column 2 = A001477. Column 3 = A005408(k). Column 4 = A016789(k+1). Column 5 = A004767(k+2). Column 6 = A016897(k+3). Column 7 = A016969(k+4). Column 8 = A017053(k+5). Column 9 = A004771(k+6). Column 10 = A017257(k+7).

Cf. A000012, A001477, A004767, A004771, A005408, A016789, A016897, A016969, A017053, A028387, A067998, A126719.

A := proc(k,n) (n+k-2)*(n-1)-1 ; end: for d from 1 to 13 do for n from 1 to d do printf("%d,",A(d-n+1,n)) ; od: od: # R. J. Mathar, Jul 10 2009

a[n_, k_] := a[n, k] = n*k - (n + k); ColumnForm[Table[a[n, k], {n, 10}, {k, n}], Center] (* Alonso del Arte, Jul 09 2009 *)

A[k,n] = (n+k-2)*(n-1) - 1.

Antidiagonal sum: Sum_{n=1..d} A(d-n+1,n) = d*(d^2-2d-1)/2 = -A110427(d). - R. J. Mathar, Jul 10 2009

Duplicate of 6th antidiagonal removed by R. J. Mathar, Jul 10 2009
Keyword:tabl added by R. J. Mathar, Jul 23 2009
Edited by N. J. A. Sloane, Sep 14 2009. There was a comment that the defining formula was wrong, but it is perfectly correct.

A257645 a(n) = 15*n + 14.

Original entry on oeis.org

14, 29, 44, 59, 74, 89, 104, 119, 134, 149, 164, 179, 194, 209, 224, 239, 254, 269, 284, 299, 314, 329, 344, 359, 374, 389, 404, 419, 434, 449, 464, 479, 494, 509, 524, 539, 554, 569, 584, 599, 614, 629, 644, 659, 674, 689, 704, 719, 734, 749, 764, 779

Offset: 0

Arkadiusz Wesolowski, Nov 05 2015

A123159(a(n)) <= 4.
This is not a subsequence of A047725 (for example, 239 is missing in A047725). - Bruno Berselli, Nov 06 2015
Equivalently, intersection of A016897 and A016789. - Bruno Berselli, Jan 24 2018

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A008597, A016897, A175887, A123159, A204542.

Magma
```
[15*n+14: n in [0..51]];
```
Maple
```
seq(15*n+14, n=0..51);
```
Mathematica
```
15 Range[50] - 1
```
PARI
```
for(n=0, 51, print1(15*n+14, ", "));
```

G.f.: (14 + x)/(1 - x)^2.
a(n) = A008597(n+1) - 1. - Omar E. Pol, Nov 05 2015
a(n) = A016897(3n+2) = A175887(2n+2) = A204542(4n+4). - Bruno Berselli, Nov 06 2015
E.g.f.: (15*x + 14)*exp(x). - G. C. Greubel, Apr 23 2018
a(n) = 2*a(n-1)-a(n-2). - Wesley Ivan Hurt, Dec 27 2023

A266297 Numbers whose last digit is a square.

Original entry on oeis.org

0, 1, 4, 9, 10, 11, 14, 19, 20, 21, 24, 29, 30, 31, 34, 39, 40, 41, 44, 49, 50, 51, 54, 59, 60, 61, 64, 69, 70, 71, 74, 79, 80, 81, 84, 89, 90, 91, 94, 99, 100, 101, 104, 109, 110, 111, 114, 119, 120, 121, 124, 129, 130, 131, 134, 139, 140, 141, 144, 149

Offset: 1

Wesley Ivan Hurt, Dec 26 2015

Numbers ending in 0, 1, 4 and 9.
Union of A008592, A017281, A017317 and A017377. - Hurt
None of these numbers are prime in Z[phi] (where phi = 1/2 + sqrt(5)/2 is the golden ratio), since the numbers in this sequence that are prime in Z can be expressed in the form (a - b sqrt(5))(a + b sqrt(5)). - Alonso del Arte, Dec 30 2015
Union of A197652 and A016897. - Wesley Ivan Hurt, Dec 31 2015
Union of A146763 and A090771. - Wesley Ivan Hurt, Jan 01 2016

G. C. Greubel, Table of n, a(n) for n = 1..5000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A008592, A016897, A017281, A017317, A017377, A090771, A091084, A141158, A146763, A197652.

[(10*n-11+(-1)^n+(4+2*(-1)^n)*(-1)^((2*n-1+(-1)^n) div 4))/4: n in [1..60]]; // Vincenzo Librandi, Dec 27 2015

A266297:=n->(10*n-11+(-1)^n+(4+2*(-1)^n)*(-1)^((2*n-1+(-1)^n)/4))/4: seq(A266297(n), n=1..100);

Table[(10 n - 11 + (-1)^n + (4 + 2 (-1)^n)*(-1)^((2 n - 1 + (-1)^n)/4))/4, {n, 50}] (* G. C. Greubel, Dec 27 2015 *)
LinearRecurrence[{1, 0, 0, 1, -1}, {0, 1, 4, 9, 10}, 60] (* Vincenzo Librandi, Dec 27 2015 *)
CoefficientList[Series[x*(1 + 3*x + 5*x^2 + x^3)/((x - 1)^2*(1 + x + x^2 + x^3)), {x, 0, 100}], x] (* Wesley Ivan Hurt, Dec 30 2015 *)
Flatten[Table[10n + {0, 1, 4, 9}, {n, 0, 19}]] (* Alonso del Arte, Dec 30 2015 *)
Select[Range[0,150],MemberQ[{0,1,4,9},Mod[#,10]]&] (* Harvey P. Dale, Jul 30 2019 *)

is(n) = issquare(n%10); \\ Altug Alkan, Dec 29 2015

G.f.: x^2*(1 + 3*x + 5*x^2 + x^3)/((x - 1)^2*(1 + x + x^2 + x^3)).
a(n) = a(n - 1) + (n - 4) - a(n - 5) for n > 5.
a(n) = (10n - 11 + (-1)^n + (4 + 2(-1)^n) * (-1)^((2n - 1 + (-1)^n)/4))/4.
a(n+1) - a(n) = A091084(n+1) for n>0.
Sum_{n>=2} (-1)^n/a(n) = (14*sqrt(5)*arccoth(sqrt(5)) - 2*Pi*sqrt(1-2/sqrt(5)) + 16*log(2) + 5*log(5))/40. - Amiram Eldar, Jul 30 2024

A215205 a(n) = (-1)^n * (A060819(n) + A060819(n+1)).

Original entry on oeis.org

1, -2, 4, -4, 6, -8, 10, -9, 11, -14, 16, -14, 16, -20, 22, -19, 21, -26, 28, -24, 26, -32, 34, -29, 31, -38, 40, -34, 36, -44, 46, -39, 41, -50, 52, -44, 46, -56, 58, -49, 51, -62, 64, -54, 56, -68, 70, -59, 61, -74, 76, -64, 66, -80, 82, -69, 71, -86, 88, -74, 76, -92, 94, -79, 81, -98, 100, -84

Offset: 0

Paul Curtz, Aug 06 2012

a(-1)=1=a(0).
a(n) - a(n-1) = b(n) = 0, -3, 6, -8, 10, -14, 18, -19, 20, -25, 30, -30, 30, -36, 42, -41, ... .
Missing terms in abs(a(n)):
PIII(n) = 0, 3, 5, 7, 12, 13, 15, 17, 18, 23, 25, 27, 30, 33, 35, 37, 42, ... . See A063241(n+1) and 6*A047222(n+1).
Quasipolynomial of order 4. - Charles R Greathouse IV, Aug 06 2012

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

Cf. A016861, A016933, A016957, A016897.

a[n_] := Switch[Mod[n, 4], 0, 5n/4+1, 1, (-3n-1)/2, 2, 3n/2+1, 3, (-5n-1)/4]; Table[a[n], {n, 0, 67}] (* Jean-François Alcover, Nov 08 2012 *)

a(4*n) = 1+5*n, a(1+4*n) = -2-6*n, a(2+4*n) = 4+6*n, a(3+4*n) = -4-5*n.
a(n+4) - a(n) = period of length 4: repeat 5,-6, 6, -5.
a(n) = 2*a(n-4) + a(n-8).
G.f. ( -1+x-3*x^2-3*x^4+x^3+x^5-x^6 ) / ( (x-1)*(1+x)^2*(x^2+1)^2 ). - R. J. Mathar, Aug 07 2012
a(n) = (5+(2*n+1)*(11*(-1)^n-(-1)^((2*n-1+(-1)^n)/4))+(-1)^((6*n-1 +(-1)^n)/4))/16. - Luce ETIENNE, Jun 05 2015

cp's OEIS Frontend

A120433 Numbers whose Roman numeral representation uses the subtractive notation.

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A144204 Array A(k,n) = (n+k-2)*(n-1) - 1 (k >= 1, n >= 1) read by antidiagonals.

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A257645 a(n) = 15*n + 14.

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A266297 Numbers whose last digit is a square.

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A215205 a(n) = (-1)^n * (A060819(n) + A060819(n+1)).

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